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Source: IMO 2022 Problem 3

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mathisreaI

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mathisreaI

#1 h Jul 13, 2022, 2:53 AM • 5 Y

Y by S.Ragnork1729, Honestcobra, cubres, farhad.fritl, Kingsbane2139

Let $k$ be a positive integer and let $S$ be a finite set of odd prime numbers. Prove that there is at most one way (up to rotation and reflection) to place the elements of $S$ around the circle such that the product of any two neighbors is of the form $x^2+x+k$ for some positive integer $x$ .

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lminsl

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lminsl

#2 h Jul 13, 2022, 2:57 AM • 11 Y

Y by AnonymousBunny, AndreiVila, MassiveMonster, lkjin7932, hitlers3x45333563, Rounak_iitr, hdnlz, bhan2025, cubres, farhad.fritl, Kingsbane2139

We solve this strengthened version of the problem:

Quote:

Let $k$ be a positive integer and let $S$ be a finite set of odd prime numbers. Prove that there is at most one
way (up to rotation and reflection) to place the elements of $S$ around a circle such that the product of any two
neighbours is of the form $x^2 + x + k$ for some integer $x$ .

We induct on $|S|$ . For $|S| \le 3$ there is at most one way to place the primes regardless, so the statement follows.

Now suppose the statement holds for all sets $S$ with, say, $|S|<n$ . The main idea is the following:

Claim. Let $p$ be the largest prime in $S$ . Then there is exactly one way to determine the neighbors of $p$ . Furthermore, if $p$ is adjacent to $q$ and $r$ , then the product $qr$ is also of the form $x^2+x+k$ .

It is clear that this claim kills the problem; say if there are two different possible configurations, then we can remove $p$ from the circle and use the induction hypothesis on $S \setminus \{p\}$ . Hence it suffices to prove the claim.

Write $pq = a^2+a+k$ and $pr = b^2+b+k$ for some nonnegative integers $a$ and $b$ . Then since
$p^2 > pq = a^2+a+k > a^2$ we have $0 \le a < p$ and similarly we get $0 \le b < p$ . Then $a$ and $b$ are two different solutions to the equation $x^2+x+k \equiv 0 \pmod p$ within $[0, p-1]$ , so $a$ and $b$ are uniquely determined by Lagrange's theorem.

Furthermore, since $a+b \equiv -1 \pmod p$ by Vieta's formula and $1<a+b < 2p-1$ , we must have $a+b=p-1$ .
Therefore, for $x = 2a+1, y=2b+1, s= 4k-1$ we have
$\begin{align*} p^2qr = (a^2+a+k)(b^2+b+k) &= \frac{1}{16} (x^2+s)(y^2+s) \\ &= \frac{1}{16} \left[ (xy-s)^2+ s(x+y)^2 \right] \\ &= \frac{1}{16} \left[ (xy-s)^2+ 4s(a+b+1)^2 \right] \\ &= \frac{1}{16} \left[ (xy-s)^2+ 4sp^2 \right] \\ &= \frac{1}{4} \left[((xy-s)/p)^2+ s \right]p^2, \end{align*}$ so for the integer $u$ satisfying $(xy-s)/p=2u+1$ we have $qr = u^2+u+k$ . We’re done.

This post has been edited 2 times. Last edited by lminsl, Jul 17, 2022, 5:06 PM

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hyay

181 posts

hyay

#3 h Jul 13, 2022, 3:13 AM • 7 Y

Y by adj0109, Kingsbane2139, Jalil_Huseynov, somebodyyouusedtoknow, Nartku, cubres, farhad.fritl

Allow $x$ in $x^2 + x + k$ to be zero. We start by proving the following lemma.

Lemma. If $p, q, r$ are odd primes and there exist $x, y \in \mathbb{Z}_{\geq 0}$ satisfying
$p = x + y + 1, \quad pq = x^2 + x + k, \quad pr = y^2 + y + k$ then there exists $z \in \mathbb{Z}_{\geq 0}$ satisfying
$qr = z^2 + z + k.$
Proof. Rewrite as
$\begin{align*} 4pq &= (2x + 1)^2 + (4k - 1) = (2x + 1 + \sqrt{1 - 4k})(2x + 1 - \sqrt{1 - 4k}) \\ 4pr &= (2y + 1)^2 + (4k - 1) = (2y + 1 + \sqrt{1 - 4k})(2y + 1 - \sqrt{1 - 4k}) \\ \end{align*}$ then
$16p^2qr = (4xy + 2x + 2y + 2 - 4k + (2x + 2y + 2)\sqrt{1 - 4k})(4xy + 2x + 2y + 2 - 4k - (2x + 2y + 2)\sqrt{1 - 4k})$ so
$4qr = \left( \frac{2xy + p - 2k}{p} + \sqrt{1-4k} \right) \left( \frac{2xy + p - 2k}{p} - \sqrt{1-4k} \right)$ meanwhile
$2xy + p - 2k = 2x(p - x - 1) + p - 2k = p(2x + 1) - 2(x^2 + x + k) = p(2x - 2q + 1)$ thus
$4qr = \left( 2x - 2q + 1 + \sqrt{1-4k} \right) \left( 2x - 2q + 1 - \sqrt{1-4k} \right)$ and so take $z = \frac{|2x - 2q + 1| - 1}{2}$ . $\blacksquare$

We will now solve the problem by induction on the number of elements of $S$ . For $|S| = 2$ , there is nothing to prove. Assume that for $|S| = k \ge 2$ the statement is true. Consider $|S| = k+1$ and take the largest prime $p \in S$ . Let $q, r < p$ be its neighbors.

Let $x, y \in \mathbb{Z}_{\geq 0}$ such that
$pq = x^2 + x + k, \quad pr = y^2 + y + k$ then
$p | x^2 + x - y^2 - y = (x - y)(x + y + 1)$ If $p | x - y$ , then one of $x, y$ is at least $p$ , but $x^2 + x + k$ and $y^2 + y + k$ are at most $p^2$ , contradiction.
So, $p | x + y + 1$ , and if $x + y + 1 \geq 2p$ then similarly we arrive at the same contradiction. Thus, $p = x + y + 1$ . Hence, if we remove $p$ from the circle and let $q$ be adjacent to $r$ , the resulting circle with $k$ primes will also satisfy the condition, due to our lemma.

There is at most one such arrangement, and considering $p | (2x + 1)^2 + (4k - 1)$ , $x, y < p$ , we know there is exactly one possible value of the pair $(x, y)$ up to permutation. So, the $k+1$ primes must also have at most one possible arrangement.

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IndoMathXdZ

694 posts

IndoMathXdZ

#4 h Jul 13, 2022, 3:21 AM • 3 Y

Y by Rounak_iitr, pavel kozlov, cubres

2022 IMO/3 wrote:

Let $k$ be a positive integer and let $S$ be a finite set of odd prime numbers. Prove that there is at most one way (up to rotation and reflection) to place the elements of $S$ around a circle such that the product of any two neighbours is of the form $x^2 + x + k$ for some positive integer $x$ .

We'll prove this by induction on $|S|$ on the relaxed problem: we'll prove the statement such that product of any two neighbors is of the form $x^2 + x + k$ for some integer $x$ , which proves the original problem. The result is clearly true for $|S| \le 3$ . Now, assume that the result is true for any $|S| < n$ , where $n \ge 4$ . We'll prove that this is true for $|S| = n$ . Indeed, suppose that there exists a configuration for $S$ with $|S| = n$ , otherwise we are done. Call a number $m$ of the form $m = x^2 + x + k$ for some $x \in \mathbb{Z}$ as nice. Now, take the largest prime $P$ on this set $S$ . By assumption, there exists distinct primes $Q, R < P$ such that
$PQ = x^2 + x + k \ \text{and} \ PR = y^2 + y + k$ Claim 01. $QR$ is nice.
Proof. By maximality of $P$ , we have $x, y < P$ and furthermore $P \mid x^2 - y^2 + x - y = (x - y)(x + y + 1) \implies x \equiv y \pmod{P} \ \text{or} \ x + y + 1 \equiv 0 \pmod{P}$ As $x \not= y$ , we conclude that $x + y = P - 1$ . Now, we have
$2(k - xy) \equiv (x + y)^2 - 2xy + (x + y) + 2k \equiv (x^2 + x + k) + (y^2 + y + k) \equiv 0 \pmod{P}$ and since $P$ is odd, then $xy \equiv k \pmod{P}$ . Write $xy = k + Pz$ for some $z \in \mathbb{Z}$ .
Therefore, we have
$\begin{align*} QR &= \frac{(PQ)(PR)}{P^2} \\ &= \frac{(x^2 + x + k)(y^2 + y + k)}{P^2} = \frac{(x^2 + x + xy - Pz)(y^2 + y + xy - Pz)}{P^2} = \frac{(Px - Pz)(Py - Pz)}{P^2} \\ &= (x - z)(y - z) = xy - z(x + y) + z^2 = k + Pz - z(x + y) + z^2 = k + z + z^2 \end{align*}$ We thus conclude that $QR$ has to be nice as well.

The above claim implies that we can construct a valid configuration with $|S| = n - 1$ assuming that a valid configuration with $|S| = n$ exists; and by inductive hypothesis, there exists a unique valid configuration for $|S| = n - 1$ , by deleting the largest prime. Therefore, the valid configuration for $|S| = n$ can only be obtained by inserting $P$ between any two consecutive primes in the configuration for $|S| = n - 1$ . Now, suppose there are more than one ways to insert the prime, then there exists distinct $a,b,c \in S$ and $x,y,z \in \mathbb{N}$ such that
$\begin{align*} aP &= x^2 + x + k \\ bP &= y^2 + y + k \\ cP &= z^2 + z + k \end{align*}$ However, as $a,b,c < P$ , we have $x,y,z < P$ as well. Furthermore, as $a,b,c$ are distinct, then $x,y,z$ are distinct as well. Therefore, $P \mid x + y + 1, y + z + 1 \implies x + y = P - 1 = y + z \implies x = z$ , a contradiction. Therefore, the configuration for $|S| = n$ has to be unique as well, and hence our original result holds.

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juckter

323 posts

juckter

#5 h Jul 13, 2022, 4:49 AM • 1 Y

Y by cubres

So before I actually write my solution, WTF was the point of making $x$ a positive integer instead of a non-negative integer? Do the leaders think it's cool to punish contestants that figure out the mathematical content of the problem, but don't notice some silly calculation quirk? Is it even cooler when someone tries to fix it by making them integers, but then applies a size argument that is technically wrong? Do they find it absolutely hilarious to hand out $6$ s instead of $7$ s? It seems so pointless to have a troll like that in what is otherwise probably the best or second-best problem this IMO.

We prove the more general version of the problem where $x$ is allowed to be a non-negative integer (*sighs*). We proceed by induction on $|S|$ where the cases with $|S| \le 3$ are immediate. Assume that $S$ admits at least one circular list as described in the problem, let $p = \max S$ and let $q$ and $r$ be its neighbors. Thus there exist non-negative integers $x$ and $y$ such that

$\begin{align*} pq &= x^2 + x + k \\ pr &= y^2 + y + k \end{align*}$
Since $p = \max S$ we have $pq < p^2$ and therefore we must have $x < p$ . Similarly $y < p$ . Subtracting, we obtain $p(q - r) = x^2 + x - y^2 - y = (x - y)(x + y + 1)$ , and as $p$ is prime it must either divide $x - y$ or $x + y + 1$ . The former gives $|x - y| \ge p$ since $q \neq r$ implies $x\neq y$ , but this is impossible for $0 \le x, y < p$ . Therefore $p$ dividess $x + y + 1$ , and since $0 < x + y + 1 \le 2(p - 1) + 1 < 2p$ it follows that $x + y + 1 = p$ . Remembering that $p(q - r) = (x - y)(x + y + 1)$ we obtain

$x - y = q - r, \qquad x + y = p - 1$
And therefore

$\boxed{x = \frac{p + q - r - 1}{2} \qquad y = \frac{p - q + r - 1}{2}}$
Now if some $s \in S$ with $s \neq q, r$ satisfied $ps = z^2 + z + k$ we would similarly be able to conclude $x = \frac{p + q - s - 1}{2}$ but this is a clear contradiction to $r \neq s$ . Therefore no other $s \in S$ satisfies this condition, and thus $q$ and $r$ must be the neighbors of $p$ in any suitable list formed by the elements of $S$ . We now substitute $x$ back into $pq = x^2 + x + k$ and after some calculations we obtain

$k = \frac{2(pq + qr + rp) - (p^2 + q^2 + r^2) + 1}{2}$
Our goal is now to show that $qr$ is also of the form $z^2 + z + k$ for some non-negative $z$ . Indeed, we can verify that $z = \frac{-p + q + r - 1}{2}$ satisfies $z^2 + z + k = qr$ . (This is to be expected as it is just relabeling $(p, q, r) \to (q, r, p)$ in $x^2 + x + k = qr$ , and $k$ is symmetric in $p$ , $q$ and $r$ ). This is suitable as long as $q + r \ge p + 1$ , otherwise if $q + r \le p - 1$ we may choose $z = \frac{p - q - r - 1}{2}$ , which provides the same value of $z^2 + z$ . The remaining case $q + r = p$ is not possible since $p, q, r$ are all odd. (This is the only thing we need their oddness for!).

Finally the previous observation implies that if we delete $p$ from any circular list, the resulting list still satisfies the conditions of the problem. By induction there is a at most one way to create a valid circular list from $S \setminus \{p\}$ , and since $p$ is forced to be between $q$ and $r$ there is also at most one way to create a valid list from $S$ .

Remark. In my opinion, exploring the case $|S| = 3$ is extremely informative -- you can prove that it is always possible to choose a suitable $k$ , and that it in fact only depends on the expressions for $pq$ and $pr$ , after which the solution flows very naturally.

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Tintarn

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Tintarn

#6 h Jul 13, 2022, 5:27 AM • 5 Y

Y by WallyWalrus, LinearAlgebra, IAmTheHazard, pavel kozlov, Assassino9931

Perhaps the shortest way to prove the required identity for $qr$ is the following:
As before, if $p>q,r$ and $pq=x^2+x+k$ and $pr=y^2+y+k$ , then $x+y=p-1$ and $x-y=q-r$ .
Hence we can write $x=q+d, y=r+d$ so that $p=q+r+2d+1$ and plugging in, we get
$q(q+r+2d+1)=(q+d)^2+q+d+k$ which immediately simplifies to $qr=d^2+d+k$ .

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Tintarn

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Tintarn

#7 h Jul 13, 2022, 5:39 AM

Y by

juckter wrote:

From what I have heard, the jury (as well as the PSC) did not realize this issue until it was too late to change the problem statement anymore.
So the problem was stated like this, but it was decided that no point will be deducted for missing out on the positivity.
Still, it is an unfortunate situation...

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juckter

323 posts

juckter

#8 h Jul 13, 2022, 5:44 AM • 2 Y

Y by Mango247, Mango247

Tintarn wrote:

juckter wrote:

Yeah, I have just learned that the proposer was also unaware of this, so in the end it was just an unintended result of losing three coin flips in a row... Still, other than that little detail the problem is very good! Definitely better to have it like this than to not have it at all.

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JG666

287 posts

JG666

#9 h Jul 13, 2022, 6:06 AM • 3 Y

Y by David-Vieta, HamstPan38825, straight

Here is a nice solution found by Chen Chen (陈晨), tutor at Xueersi School in China.

Solution. We induct on $|S|=n$ , with base case $(n=2)$ being clear. Now suppose the statement holds for all $|S|\leqslant n$ and we check $n+1$ :

Set $S=\{p_1<p_2<\cdots<p_{n+1}\}$ . First we will show that the two numbers which $p_{n+1}$ is connected to are uniquely determined.

Let $p_ip_j = x^2+x+k = (2x+1)^2+(4k-1)$ . Set $4k-1=t$ for convenience. Also, let
$\begin{cases} 4p_ip_{n+1} = a^2+t\quad (1) \\ 4p_jp_{n+1} = b^2+t\quad (2) \\ \end{cases}$ where $a,b\in\mathbb {N_+}$ . Notice that $a,b$ are both odd, $a,b<2p_{n+1}$ , $p_i, p_j < p_{n+1}$ . By subtracting $(2)-(1)$ we get:
$4(p_j-p_i)p_{n+1} = (b+a)(b-a)$ which implies $b+a = 2p_{n+1}$ and $b-a = 2(p_j-p_i)$ .

Since $a,b$ are uniquely determined by $t$ , we must have $p_i$ and $p_j$ uniquely determined as well.

To finish, note
$\begin{align*} 4p_ip_j &= \frac{(a^2+t)(b^2+t)}{4p_{n+1}^2} \\ &= \frac{\left[(ab-t)^2+t(a+b)^2\right]}{4_{n+1}^2} \\ & = \left(\frac{ab-t}{2p_{n+1}}\right)^2+t \end{align*}$ Then cut out $p_{n+1}$ and induct down. Done.

Remarks by JG666: Induction is a intuitive idea. I think the most amazing part is that he used $p_ip_j = x^2+x+k = (2x+1)^2+(4k-1)$ , which I didn't come up with. After we had this then the left process is quite natural.

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rama1728

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rama1728

#10 h Jul 13, 2022, 6:29 AM • 2 Y

Y by rightways, The_Great_Learner

Extraordinary problem. First time I am seeing such a good NT in IMO P3/6.

There are two parts in this problem. The first part is to prove that we can actually induct, which means, the flow of the inductive step is guaranteed. The second part is to actually complete the induction.

Part 1. Guaranteeing Induction.

Let $P$ be the largest prime in $S$ . I claim that $P$ 's neighbors $q$ and $r$ satisfy the following: $qr=x^2+x+k$ for some $x$ . Note that if we prove this, we are done because we can delete $P$ and work on the inductive step.

Assume that $qP=a^2+a+k$ and $rP=b^2+b+k$ Subtracting gives: $P\mid (a-b)(a+b+1)$ Assume that $p\mid a-b$ . Then, we have $a$ or $b$ to be greater than $P$ , assume that $a>P$ . Then, $P^2>qP=a^2+a+k>P^2$ a contradiction. Therefore, $P\mid a+b+1$ . Now, let $a+b+1=Px$ Then, $\begin{align*} 2P^2 &> 2P(q+r) \\ &= (a^2+a+k)+(b^2+b+k) \\ & = (a+b)^2+(a+b)+k-2ab \\ &= (Px-1)^2+(Px-1)+k-2ab \\ &> (Px-1)^2+(Px-1)-\frac{(Px-1)^2}{2}+2k &= \frac{(Px)^2}{2}+2k-\frac{1}{2} \\ \end{align*}$ This evidently implies $x=1$ , because if $x>1$ , the inequality fails. Therefore $a+b+1=P$ . Now, that means $(a+b+1)q=a^2+a+k$ and $(a+b+1)r=b^2+b+k$ so $a+b+1\mid a^2+a+k$ which then gives $a+b+1\mid ab-k$ . Let $d=\frac{ab-k}{a+b+1}$ then we have $q=a-d$ and $r=b-d$ . So, $qr=(d-a)(d-b)=d^2-(a+b)d+ab=d^2-(P-1)d+Pd+k=d^2+d+k$ and we are done.

Part 2. Completing the inductive step.

Now, we can induct. Assume that the statement holds for $\lvert S\rvert=n$ . Then, if we add a bigger prime $P$ to the set, it will be placed between two primes $q$ and $r$ . Moreover, since the number of solutions to $x^2+x+k\equiv0\pmod{P}$ is precisely $2$ , $P$ 's neighbors $q$ and $r$ are uniquely determined, completing the inductive step and thus solving the problem. $\blacksquare$

In my opinion, this problem is a pinnacle of number theory. I hope more such problems come up in the future!

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VicKmath7

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VicKmath7

#11 h Jul 13, 2022, 6:35 AM

Y by

Probably this is not the cleanest solution, but I also used the same idea of inducting by removing the largest prime, and the difference is that I am using the existence of $x$ in Case 1 with discriminants. Even though, this is remarkably straightforward for P3, I just tried induction and it worked. Here it goes:

Before we begin, we consider some basic facts about an equation $x^2+x+k=pq$ . In order to have an positive integer root, we need $D=1-4(k-pq)=4pq+1-4k=4pq-m$ to be a perfect square (let $4k-1=m$ ); note that this is also a sufficient condition.

We prove the statement by induction on the number of elements in $S$ , the base cases are trivial. Assume it is true for $k-1$ , we will prove it for $k$ . Consider primes $p_1, p_2,...,p_k$ and let WLOG $p_k$ be the largest prime among them. Suppose there are two working arrangements for these primes. Let's remove $p_k$ , and now we obtain either two identical working arrangements for $p_1,...,p_{k-1}$ , or at least one arrangement which does not work for $p_1,...,p_{k-1}$ (two distinct working arrangements are forbidden due to IH).

Case 1. We have one non-working arrangement, but it becomes working after adding $p_k$ . Hence there is exactly one pair $(p_i,p_j)$ (let WLOG $p_i>p_j$ ) in that arrangement that doesn't have a product of the desired type, but $(p_i,p_k)$ and $(p_j,p_k)$ do have such a product. Hence we have $4p_ip_k-m=a^2, 4p_jp_k-m=b^2$ and $a>b$ . Apparently $a,b<2p_k$ since $p_k>p_i,p_j$ . We have that $4p_k(p_i-p_j)=(a-b)(a+b)$ . We have that $a-b<2p_k$ , so if $p_k$ divides it, then $a-b=p_k$ is odd and $a+b=4(p_i-p_j)$ is even, absurd. Hence $p_k$ divides $a+b<4p_k$ , and similarly $a+b$ can't be odd, so $a+b=2p_k, a-b=2(p_i-p_j)$ , hence $a=p_k+p_i-p_j,b=p_k+p_j-p_i$ , so plugging back in the equations, we obtain $m=2p_ip_k+2p_jp_k+2p_ip_j-p_i^2-p_j^2-p_k^2$ . Now, note that $4p_ip_j-m=(p_i+p_j-p_k)^2$ , hence $p_ip_j=x^2+x+k$ has a positive integer root, contradiction.

Case 2. We obtain two identical working arrangements after removing $p_k$ . We will prove that for a given working arrangement, we can add $p_k$ in exactly one place, which will finish. So let's assume $p_k$ can be added between $(p_i,p_j)$ , and $(p_r, p_s)$ , WLOG $p_i>p_j, p_r>p_s$ . As in the first case, we obtain $m=2p_ip_k+2p_jp_k+2p_ip_j-p_i^2-p_j^2-p_k^2=2p_rp_k+2p_sp_k+2p_rp_s-p_r^2-p_s^2-p_k^2$ , so $2p_k(p_i+p_j)-(p_i-p_j)^2=2p_k(p_r+p_s)-(p_r-p_s)^2$ , which can be rewritten as $2p_k(p_i+p_j-p_r-p_s)=(p_i-p_j-p_r+p_s)(p_i-p_j+p_r-p_s)$ . Using that $p_k$ is the biggest prime, we obtain that if WLOG $p_i-p_j>p_r-p_s$ , then both expressions on the RHS are smaller than $2p_k$ but both are even, so $p_k$ can't divide neither of them. Now suppose that some of those two terms are zero. Then we have $p_j=p_r$ or $p_j=p_s$ , so we obtain a quadruple of primes $(p, q, r, s)$ , such that putting $p>q, r, s$ between $(q, r)$ and $(r, s)$ again gives working configuration, so any pair of those 4 primes has product of the type $x^2+x+k$ . Let $pq=x^2+x+k, pr=y^2+y+k, ps=z^2+z+k$ . Obviously $x, y, z<p$ , so since $p|(x-y)(x+y+1)$ we have $p|x+y+1<2p$ , hence $p=x+y+1=y+z+1=z+x+1$ , so $x=y=z$ , absurd.

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Pmshw

17 posts

Pmshw

#12 h Jul 13, 2022, 7:21 AM • 1 Y

Y by alinazarboland

Here's a sketch of the solution
Lemma.Let $p_{i-1} \le p_{i+1}$ where $p_{i-1},p_{i},p_{i+1}$ are three consecutive primes on the circle. Then we have : $p_{i-1}+p_{i} \leq p_{i+1}$ for all elements of $S$ except 1 of them.
Proof . Let $f(x_i)=p_i.p_{i+1}$ . then we have either $x_i=x_{i+1} (mod p)$ or their sum is -1 mod p. In first case , by taking the difference $f(x_{i+1})-f(x_i) \geq f(x_i +p)-f(x_i) \geq p_i^2$ we'll get the desired inequality. In the second case , let $x_i=pl+r$ where $r$ is between 0,p-1 . by adding up the relations for $x_i , x_i+1$ we'll get the fact $l$ cannot be 1 by taking mod 2. and $l \geq 2$ . Let $x_{i+1}=tp+p-r-1$ . if $t\geq l+1$ , we're done by the same inequalities as before and there is at most one index $i$ st $t=l$ and we have $t \geq l$ and we're done.
Now , assume wlog $p_3>p_2+p_1$ . So $p_3>p_2$ and by the lemma $p_4>p_3+p_2$ and so on...and clearly there is only one place we can stop in this proggress and we're done.

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Justanaccount

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Justanaccount

#13 h Jul 13, 2022, 8:06 AM • 3 Y

Y by JG666, Mango247, Mango247

My sketchy solution:

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ltf0501

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ltf0501

#14 h Jul 13, 2022, 8:53 AM • 1 Y

Y by rama1728

Tintarn wrote:

juckter wrote:

That's right. My colleague and I discovered this issue when we were translating this problem. But it was too late to modify the English version, so the problem remained the same constraint positive integer $x$ . At least it was found before the exam, or it might be embarrassed if we didn't discover this problem and found by the contestants. Also, it might be a serious issue when coordinating

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rama1728

800 posts

rama1728

#15 h Jul 13, 2022, 9:00 AM

Y by

ltf0501 wrote:

Tintarn wrote:

juckter wrote:

That's right. My colleague and I discovered this problem when we were translating this problem. But it was too late to modify the English version, so the problem remained the same constraint positive integer $x$ . At least it was found before the exam, or it might be embarrassed if we didn't discover this problem and found by the contestants. Also, it might be a serious issue when coordinating

WOW YOU MADE THIS PROBLEM!!! IMO 2020 P6 AND IMO 2022 P3 WOW MAN! KEEP GOING!

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ltf0501

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ltf0501

#16 h Jul 13, 2022, 9:17 AM • 4 Y

Y by Mango247, Mango247, Mango247, EpicBird08

rama1728 wrote:

WOW YOU MADE THIS PROBLEM!!! IMO 2020 P6 AND IMO 2022 P3 WOW MAN! KEEP GOING!

Oh sorry, I mislead you lol. This problem is not proposed by me. The issue I found is that the official solution and the whole jury did not notice the constraint positive integers $x$ leads to some strange issues. We just tried to reflect on this issue at the jury meeting.

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rama1728

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rama1728

#17 h Jul 13, 2022, 10:07 AM

Y by

ltf0501 wrote:

rama1728 wrote:

WOW YOU MADE THIS PROBLEM!!! IMO 2020 P6 AND IMO 2022 P3 WOW MAN! KEEP GOING!

Ohh.. I see, thanks!

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math90

1477 posts

math90

#18 h Jul 13, 2022, 12:09 PM

Y by

We solve this strengthened version of the problem:

Quote:

We induct on $|S|$ . For $|S| \le 3$ there is at most one way to place the primes regardless, so the statement follows.

Now suppose $n>3$ and the statement holds for all sets $S$ with $|S|=n-1$ . For an odd prime $p$ , we say that an odd prime $q$ is $p$ -good if $q<p$ and there exists an integer $x$ such that $pq=x^2+x+k$ .

Claim 1. Let $p$ be an odd prime. Let $q,r$ be distinct $p$ -good primes. Then
$p^2+q^2+r^2-2pq-2pr-2qr=1-4k.$
Proof. WLOG assume $q>r$ . Then there exist integers $x,y$ such that $pq=x^2+x+k$ and $pr=y^2+y+k$ . If $x<0$ we can replace $x$ by $-1-x$ , and similarly for $y$ . Hence we can assume $x,y\ge 0$ . Since $q>r$ then $x>y$ . We see that
$p(q-r)=x^2+x-y^2-y=(x-y)(x+y+1).$ Since $k$ is a positive integer,
$p^2>pq=x^2+x+k>x^2.$ Hence $x<p$ . Therefore $y<x<p$ , so $x-y<p$ . Since $p\mid (x-y)(x+y+1)$ we must have $p\mid x+y+1$ . Now
$0<x+y+1\le (p-1)+(p-1)+1=2p-1<2p$ so $x+y+1=p$ . Hence $x-y=q-r$ so $2x+1=p+q-r$ . Now
$\begin{align*} 1-4k &=(2x+1)^2-4pq\\ &=(p+q-r)^2-4pq\\ &=p^2+q^2+r^2-2pq-2pr-2qr.\;\square \end{align*}$

Claim 2. Let $p$ be an odd prime. Then:

There exist at most two $p$ -good primes, and
if $q>r$ are two $p$ -good primes then $r$ is $q$ -good.

Proof. Let us prove part by part.

Assume by contradiction that there exists $p$ -good primes $q,r,s$ such that $p>q>r>s$ . By claim 1 we obtain
$p^2+q^2+r^2-2pq-2pr-2qr=p^2+q^2+s^2-2pq-2ps-2qs.$ $\implies r^2-2pr-2qr=s^2-2ps-2qs.$ $\implies (r-s)(r+s-2p-2q)=0.$ But $r>s$ and $r+s<2p+2q$ , contradiction.
Assume $q>r$ are two $p$ -good primes. By claim 1 we obtain
$1-4k=p^2+q^2+r^2-2pq-2pr-2qr=(q+r-p)^2-4qr.$ Since $p,q,r$ are odd then $x=\frac{q+r-p-1}{2}$ is an integer. Moreover
$x^2+x+k=\frac{(2x+1)^2-1+4k}{4}=\frac{(q+r-p)^2-1+4k}{4}=qr$ so $r$ is $q$ -good. $\square$

Back to the main problem. Let $p=\max(S)$ . By claim 2, $p$ has the same pair of neighbors in all legal arrangements, and removing $p$ preserves the conditions of the problem. Hence we are done by induction hypothesis.

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adamov1

355 posts

adamov1

#19 h Jul 13, 2022, 2:01 PM • 1 Y

Y by Mango247

Call a pair of odd primes $(p, q)$ good if $pq$ can be written as $x^2+x+k$ for some $x$ . We first prove the following lemmas

Lemma 1. If $p>q>r$ are odd primes with $(p, q)$ good and $(p, r)$ good, then $(q, r)$ is good.
Proof

Lemma 2. If $p>q>r>s$ are odd primes, we cannot have all pairs among them good.
Proof

Now proceed by induction on $|S|$ . The base case $|S|=4$ follows from Lemma 2. Now suppose we have two circular arrangements of $S$ . Let $p$ be the largest element of $S$ , then note that, by Lemmas 1 and 2 combined, $p$ must have the same neighbors $q,r$ in both arrangements.

By Lemma 1, $(q, r)$ is good. Thus if we remove $p$ from both circular arrangements and connect $q$ and $r$ , we get valid circular arrangements for $S\backslash\{p\}$ . By the inductive hypothesis, these must be the same, so the original arrangements were the same, as desired.

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DVDthe1st

341 posts

DVDthe1st

#21 h Jul 13, 2022, 7:49 PM • 2 Y

Y by IAmTheHazard, pavel kozlov

There are two key steps to solving the problem: looking at the largest prime $p$ and noticing that if $pq, pr$ can be expressed as $x^2+x+k$ , then so can $qr$ . Here I'll give some motivation for the latter.

Suppose $pq = x^2 + x + k, pr = y^2 + y + k$ . If we let $\alpha$ denote the complex number such that $\alpha^2 + \alpha + k = 0$ , then note that the other root $\bar \alpha$ is the complex conjugate, satisfying $\alpha + \bar\alpha = -1, \alpha \bar\alpha = k$ . Then we have
$x^2 + x + k = (x-\alpha)(x-\bar\alpha) =: N(x-\alpha)$ where for us $N(z) := z\bar z$ . A key fact is that the function $N$ is multiplicative and in general $N(a-b\alpha) = a^2 + ab+ kb^2$ , so we expect that
$(x^2 + x + k)(y^2 + y + k) = N((x-\alpha)(y-\alpha)) = N(xy - (x+y) \alpha+ \alpha^2) = N(xy - k - (x+y+1)\alpha) = (xy-k)^2 + (xy-k)(x+y+1) + k(x+y+1)^2$ and conveniently for us we have $x+y+1 = p$ and $p \mid xy-k$ , so by setting $A= (xy-k)/p$ we do indeed get $qr = A^2 + A + k$ .

Extra motivation, for those who know a bit of algebraic number theory

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megarnie

5610 posts

megarnie

#22 h Jul 14, 2022, 10:55 AM • 1 Y

Y by I.owais

We solve the problem

Stronger IMO 2022 Problem 3 wrote:

Let $k$ be a positive integer and let $S$ be a finite set of odd prime numbers. Prove that there is at most one way (up to rotation and reflection) to place the elements of $S$ around the circle such that the product of any two neighbours is of the form $x^2+x+k$ for some integer $x$ .

This will solve the original problem.

Call a positive integer $n$ good if it can be written as $x^2 + x + k$ , where $x$ is an integer.

Call an arrangement of the elements of $S$ in a circle to be valid if the product of any two neighbors is of the form $x^2+x+k$ , where $x$ is an integer. We wish to show that $S$ has at most one valid arrangement up to rotation and reflection.

Lemma: Let $q$ and $r$ be distinct primes in $S$ , and $p$ be the largest prime in $S$ . If $pq$ and $pr$ are good, then $qr$ is good.
Proof: Let $pq = x^2 + x + k, pr = y^2 + y + k$
Since $x^2 + x + k = (-x-1)^2 + (-x-1) + k$ and $y^2 + y + k = (-y-1)^2 + (-y-1) + k,$ since at least one of $\{-x-1,x\}$ is nonnegative, there exist integers $x,y\ge0$ with $pq = x^2 + x + k, pr = y^2 + y + k$
We have $p(q-r) = pq - pr = (x-y)(x+y+1)$
If $p\mid x-y$ , then $x=y$ because $0\le x,y<p$ . This is a contradiction as $q$ and $r$ are distinct.

So $p\mid x+y+1$ .

Since $x,y<p$ , we have $x+y+1\le 2(p-1) + 1 = 2p-1$ The only positive multiple $p$ less than $2p-1$ is $p$ , so $p=x+y+1$ .

So $x-y = q-r$ and $x+y = p-1$ . This implies $2x = p+q-r-1$ .

We have $\begin{align*} (x-q)^2 + (x-q) + k \\ = x^2 - 2xq + q^2 + x-q + k \\ = x^2 + x + k - 2xq + q^2 - q \\ = x^2 + x + k - (p+q-r-1)q + q^2 - q \\ = (x^2 + x + k) - pq -q^2 +qr + q + q^2 - q \\ =(pq - pq ) + qr \\ = qr, \\ \end{align*}$ so $qr$ is good. $\square$

Now we induct on $|S|$ . If $|S|\in \{1,2,3\}$ , there is exactly one way to place the elements of $S$ around a circle up to rotation and reflection, so the result holds.

Suppose the result is true for all sets with cardinality $n-1$ .

For a set $S$ with $|S| = n$ , we have a set $S_1\subseteq S$ , such that $|S_1| = n-1$ , and $S_1$ contains all elements of $S$ except the largest prime in $S$ .

If there is at most one valid arrangement of the elements of $S$ , then we are done. Suppose FTSOC $S$ had at least two valid arrangements.

By the inductive hypothesis, there is at most one valid arrangement of the elements of $S_1$ .

By the lemma, for each valid arrangement of $S$ , deleting the largest prime in $S$ gives the unique valid arrangement of the elements of $S_1$ . Since we are assuming $S$ has a valid arrangement, $S_1$ must also have a valid arrangement.

Therefore each valid arrangement of $S$ is made by inserting the largest prime factor in $S$ , say $p$ , between two primes in the unique valid arrangement of $S_1$ .

Claim: There is at most one way to insert $p$ between two primes in the valid arrangement of $S_1$ .
Proof: Suppose there was at least two ways. Then there exist distinct primes $a,b,c<p$ , such that $ap = x^2 + x + k, bp = y^2 + y + k, cp = z^2 + z + k,$ where $x,y,z$ are nonnegative integers (since $x^2 + x + k = (-x-1)^2 + (-x-1) + k$ , and so on).

Since $ap < p^2 < p^2 + p + k,$ $bp < p^2 < p^2 + p + k,$ and $cp < p^2 < p^2 + p + k,$ we have $0\le x,y,z<p$ .

So $p(a-b) = ap - bp = (x-y)(x+y+1)$ If $p\mid x-y$ , then $x=y$ because $0\le x,y<p$ . This is a contradiction as $a$ and $b$ are distinct.

Thus, $p\mid x+y+1$ . Since $0< x+y+1\le 2(p-1) + 1 =2p-1,$ we have $p=x+y+1$ .

Similarly, $p=x+z+1$ .

This implies $y=z$ , so $b=c$ , contradiction. $\square$

Inserting $p$ between two primes in $S_1$ uniquely determines the arrangement of $S$ . Thus, there is at most $1$ valid arrangement of $S$ .

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PUjnk

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PUjnk

#23 h Jul 14, 2022, 9:32 PM

Y by

Here's my solution :

We first prove the following lemma:
\textbf{Lemma}: If p,q,r are distinct primes with $p>q$ and $p>r$ . Now suppose $k \in \mathbb{N}$ such that $\exists m,n \in \mathbb{N}_0$ satisfying $m^2+m+k=pq$ and $n^2+n+k=pr$ , then $\exists y \in \mathbb{N}_0$ such that $y^2+y+k=qr$ or $m=q$ and $n=r$ .

\begin{proof}: $m^2+m+k=pq \Longrightarrow (2m+1)^2+(4k-1)=4pq$ - (1).
Similarly, we have that $(2n+1)^2+(4k-1)=4pr$ .
Now let $d=4k-1,a=2m+1,b=2n+1$ .
So we can rewrite (1),(2) as :
$a^2+d=4pq$ - (1) and $b^2+d=4pr$ - (2).
Now consider $K=\mathbb{Q}(\sqrt{-d})$ .
We shall denote the norm of an element $\alpha$ in $K$ as $N(\alpha)$ .
Then we see that :
$N(a+\sqrt{-d})=4pq$ - (3) and $N(b+\sqrt{-d})=4pr$ - (4).
Now from (1) and (2), we have that $p \mid a^2-b^2=(a-b)(a+b) \Longrightarrow p \mid a+b$ or $p \mid a-b$ (since p is a prime).
Now suppose $p \mid a-b$ .
Note that since $q<p,r<p$ , we have that $a^2<a^2+d=4pq<4p^2 \Longrightarrow a < 2p$ .
Similarly, we have that $b<2p$ .
But now we have that $p \mid a-b$ .
Moreover since $d \equiv 3$ (mod 4), we have that $a^2 \equiv 1$ (mod 4) and so a is odd. Similarly, we also have that $b$ is odd.
Thus $2 \mid a-b$ .
It is clear that $p \neq 2$ and hence $2p \mid a-b$ (as $g(p,2)=1$ since p is a prime $\neq 2$ ).
But now since $0 \le a,b < 2p$ , we have that $|a-b|<2p$ .
Thus we must have that $a=b$ .
However this would mean that $q=r$ , a contradiction.
Thus we must have that $p \mid a+b$ .
Again we have that $a,b$ are both odd and thus $2 \mid a+b \Longrightarrow 2p \mid a+b$ .
But now $0 < a,b < 2p \Longrightarrow 0 \le a+b < 4p$ .
Thus the only possibility is $a+b=2p,a=b=0$ .
It is easy to see that $a=b=0$ gives $q=r$ , which is not possible.
Thus we must have that $a+b=2p$ .
Now note that $(a+\sqrt{-d})(b+\sqrt{-d})=(ab-d)+(a+b)\sqrt{-d}$ .
Now we see that $a \equiv - b$ (mod p) and thus $ab-d \equiv -a^2-d \equiv 0$ (mod p).
Thus $p \mid ab-d$ .

Now note that we also have that $a,b,d$ are odd. Thus $2 \mid ab-d \Longrightarrow 2p \mid ab-d \Longrightarrow \exists x \in \mathbb{Z}$ such that $ab-d=2px$ .
Now also recall that $a+b=2p$ .
Thus $(a+\sqrt{-d})(b+\sqrt{-d})=2px+2p\sqrt{-d}=2p(x+\sqrt{-d})$ .
So taking norms on both sides and using multiplicativity of norms and (3) and (4), we get that :
$(4pq)(4pr)=4p^2(x^2+d) \Longrightarrow x^2+d=4qr$ .
WLOG assume $x \ge 0$ .
Now since $d \equiv 3$ (mod 4), we have that $x^2 \equiv 1$ (mod 4) and thus x is odd. Thus $x=2y+1$ for some $y \in \mathbb{Z}$ with $y \ge 0$ .
So plugging in, we get that :
$(2y+1)^2+d=4qr \Longrightarrow y^2+y+k=pr$ , as desired.
This proves the lemma.
\end{proof}

Now back to the main problem, construct a graph G with vertices as $p_1,....,p_n$ and we say $p_i \sim p_j$ iff: $\exists x \in \mathbb{N}_0$ such that $p_i p_j=x^2+x+k$ .
So it suffices to show that $G$ has atmost one hamiltonian cycle.
We prove this by induction on n.
Base case : $n=3$ -> trivial.
Now suppose it holds for $n-1$ .
Now if G has no hamiltonian cycle, then we are trivially done.
So suppose G has a hamiltonian cycle.
WLOG we can assume that $p_1 \sim p_2 \cdots p_n \sim p_1$ .
WLOG let $p_n$ be the largest prime from $p_1,...,p_n$ .
Now first we show that $deg(p_n)=2$ .
Now we have that $p_1 \sim p_n$ and $p_{n-1} \sim p_n$ and $p_n>p_1,p_{n-1}$ .
Let $p_np_1=x^2+x+k$ and $p_np_{n-1}=y^2+y+k$ .
Then from the proof of the lemma, we know that $(2x+1)+(2y+1)=2p \Longrightarrow x+y=p-1$ .
Now suppose $\exists 1 \le j \le n-1$ such that $p_j \sim p_n$ .
Then we have that $p_np_j=z^2+z+k$ . Now if $p_j \neq p_1$ , then again by same argument, we have that $x+z=p-1=x+y \Longrightarrow y=z \Longrightarrow p_np_j=p_np_{n-1} \Longrightarrow p_j=p_{n-1}$ ie $j=n-1$
So we have that $p_n \sim p_i$ iff : $i=1$ or $i=n-1$ .
Thus $deg(p_n)=2$ .
Now from the lemma we also know that $\exists u \in \mathbb{N}_0$ such that $p_{n-1}p_1=u^2+u+k \Longrightarrow p_{n-1} \sim p_1$ .
So now consider $H=G/\{p_n\}$ .
Then since $p_1 \sim p_{n-1}$ , we have that $p_1 p_2 .... p_{n-1}$ is a hamiltonian cycle of H.
So by induction hypothesis we know that $p_1 p_2 \cdots p_{n-1}$ is the unique hamiltonian cycle of H.
Thus since $deg(p_n)=2$ , we have that $p_1 p_2 \cdots p_n$ is the unique hamiltonian cycle of $G$ .
This completes the induction step.

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renrenthehamster

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renrenthehamster

#25 h Jul 16, 2022, 1:29 PM

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I have done up a video solution for the day 1 problems and discussion of the statistics here: https://www.youtube.com/watch?v=nYD-qIOdi_c
Hopefully it is easier to understand the long-ish solution through a video. It's similar to some of the approaches above but I think we need to handle the annoying problem of "positive integer $x$ " with some care. For example in the first solution on this forum page, how do we know that $a$ and $b$ are positive if the OP decided to prove the general problem? It's not a simple typo that can just be dropped. Without $a$ and $b$ positive, now it is not clear that $a$ and $b$ are uniquely determined in the next part of the solution.

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PUjnk

71 posts

PUjnk

#26 h Jul 16, 2022, 8:40 PM

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In my proof I only need x is non-negative. So I think the generalization to "x is non-negative" is fine.

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S.Ragnork1729

215 posts

S.Ragnork1729

#27 h Jul 18, 2022, 1:04 PM

Y by

My Same Solution of the Problem : Call a pair of odd primes $(p, q)$ Favourable if $pq$ can be written as $x^2+x+k$ for some $x$ . Let us head on by proving two lemmas below .

Lemma 1. If $p>q>r$ are odd primes with $(p, q)$ Favourable and $(p, r)$ Favourable, then $(q, r)$ is Favourable .
Proof : Let $pq=x^2+x+k$ and $pr=y^2+y+k$ . Then we have
$p(q-r) = (x-y)(x+y+1)$ Note that since $p>q>r$ , we have $y<x<p$ and thus $0 < x-y < p$ and $0 < x + y + 1 < 2p$ . Since we must have $p$ dividing one of $x-y$ and $x+y+1$ , we must have $p=x+y+1$ and thus $x-y=q-r$ .

Then let $d=x-q=y-r$ . We have
$\begin{align*} d^2+d+k&=(x-q)^2+(x-q)+k\\ &=x^2-2qx+q^2+x-q+k\\ &=pq-2qx+q^2-q\\ &=q(p-2x+q-1)\\ &=q(y-x-q)\\ &=qr \end{align*}$ so $(q, r)$ is Favourable .

Lemma 2. If $p>q>r>s$ are odd primes, we cannot have all pairs among them Favourable.
Proof : Suppose otherwise. Let $pq=x^2+x+k, pr=y^2+y+k, ps=z^2+z+k$ . From the proof of the previous lemma, we have $p=x+y+1=x+z+1=y+z+1$ , so $x=y=z$ and thus $q=r=s$ , contradiction.

Proceeding by induction on $|S|$ . The trivial case $|S|=4$ follows from Lemma 2. Suppose we have two circular arrangements of $S$ . Let $p$ be the largest element of $S$ , then, by Lemmas 1 and 2 combined, $p$ must have the same neighbors $q,r$ in both arrangements.

By Lemma 1, $(q, r)$ is Favourable Thus if we remove $p$ from both circular arrangements and connect $q$ and $r$ , we get valid circular arrangements for $S\backslash\{p\}$ . By the inductive hypothesis, these must be the same, so the original arrangements were the same, as desired.

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dgrozev

2463 posts

dgrozev

#28 h Jul 27, 2022, 8:27 AM • 2 Y

Y by rafayaashary1, Rijul saini

Translated into graph theory language the problem looks like:

We have an (infinite) graph $G$ with vertices all odd prime numbers. Two vertices $p$ and $q$ are connected if and only if there exists a positive integer $x$ such that $pq=x^2+x+k.$ We want to prove that any finite subgraph $H$ of $G$ contains at most one Hamiltonian cycle.

This is a very strong claim about a graph. Its structure should not be so complicated. For instance, it holds if every vertex of $G$ has a degree at most $2$ . We can try it as a first idea, though unfortunately this is not true. Trying to prove the degrees are 2 or less, we, of course, are doomed, but we could establish that it's "almost" true. Each vertex $p$ is connected with at most two other vertices less than $p$ . Here, by "less" we mean the natural order of the primes. This is the easier part - there are only two properties of $G$ we need, this is the first one.

Lemma 1. Each vertex (prime number) $p$ is connected with at most two primes less than $p$ .

Proof. Suppose, a prime $p>2$ is connected to two primes $q,r$ with $p>q>r>2$ . Then there exist two positive integers $x,y$ such that

$pq=x^2+x+k\,;\, pr=y^2+y+k\qquad (1)$
Subtracting the two equalities, we get

$p(q-r)=(x-y)(x+y+1)$
Since $p$ is the biggest one among $p,q,r,$ by $(1)$ we obtain $x,y<p,$ thus $p$ cannot divide $x-y$ , hence it divides $x+y+1,$ but $x+y+1<2p$ therefore $x+y+1=p.$ I think, almost every contestant has discovered this fact. One step further lies the observation that $p$ cannot be connected to another prime (apart from $q,r$ ) that is less than $p$ . Indeed, if $s$ is a prime $s<p$ and $ps=z^2+z+k$ the same argument yields $x+z+1=p$ therefore $z=y$ and $s=r.$ $\blacksquare$

What we have established till now is that $G$ is (so called) 2-degenerate graph. A graph is 2-degenerate if its vertices can be enumerated as $v_1,v_2,\dots$ such that $v_{j}$ is connected to at most two vertices among $v_1,v_2,\dots,v_{j-1}$ .
Unfortunately, this property is not enough to prove the uniqueness of Hamiltonian cycles. Bellow is a picture that presents a two degenerate graph with two Hamiltonian cycles.

https://dgrozev.files.wordpress.com/2022/07/imo-2022-p3.png

Indeed, the vertices can be ordered like it's shown: 1,2,3,4,5, and every next vertex is connected with at most two of the previous ones. However, there are two Hamiltonian cycles: 1,2,3,5,4,1 and 1,3,5,4,2,1.

The next claim is the toughest NT part of this problem and is the key property $G$ satisfies.

Lemma 2. Let $p>2$ be a prime number, connected to $q$ and $r\,,\, p>q>r>2.$ Then $q$ and $r$ are also connected.

Proof. Let $pq=x^2+x+k, pr=y^2+y+k$ . As we showed $p>x>y$ and

$\displaystyle p(q-r)=(x-y)(x+y+1).$
It implies $p=x+y+1$ . Let us set $x=p-s, y=s-1$ where $1<s<p.$ Further,

$pq=p^2-2ps+p+s^2-s+k\,;\, pr=s^2-s+k\qquad (2)$
Hence, $p\mid s^2-s+k$ and let $s^2-s+k=up.$ Multiplying the equalities in $(2)$ we get

$qr=u(p-2s+1+u)=u^2-2us+u+up\qquad (3)$
Putting back in $(3)$ $up=s^2-s+k$ we obtain

$qr=(u-s)^2 +(u-s) +k$
Thus, we showed $qr=z^2+z+k$ where $z\in\mathbb{Z}$ . Here, we have a "small" issue since it may happen $z\le 0$ . If one trace back what $u$ and $s$ were, one can it's mostly the case. Fortunately, if we can represent $qr$ as $qr=z^2-z+k$ where $z> 0$ we can represent it also as $qr=(z-1)^2+(z-1)+k$ . $\blacksquare$

What we actually proved is that $qr$ can be represented as $z^2+z+k$ where $z\geq 0$ . It's slightly annoying since $q$ and $r$ are connected if and only if $qr$ is representable as $z^2+z+1$ and $z\in \mathbb{Z}^+.$ We cannot do anything, but to change the statement of the problem. Allow the vertices $p,q$ be connected even if $pq=0^2+0+k=k.$ It cannot change anything, since if we prove the claim for a graph obtained by adding a few more edges, it will be also true for the original graph.
That said, let's recap. We proved the graph $G$ has the following two properties.

(i) The vertices of $G$ can be ordered $v_1,v_2,\dots$ such that $v_j, j\ge 2$ is connected with at most two among the vertices $v_1,v_2,\dots,v_{j-1},$ and
(ii) If $v_j$ is connected to $v_{\ell}$ and $v_{m}, \ell, m<j$ then $v_{\ell}$ and $v_m$ are also connected.

This is enough to prove that there does not exist a subgraph $H$ of $G$ with two distinct Hamiltonian cycles. The argument is simple. Suppose there is a subgraph $H$ like that and let the vertex $v_m$ be the first one so that $H$ appears after adding it. That is, there is no subgraph like that in the graph $G'$ induced by $v_1,v_2,\dots,v_{m-1}$ but there is one, say $H$ , in the graph $G''$ induced by $v_1,v_2,\dots,v_{m-1},v_m.$ Apparently, $v_m\in H$ since it's the first time $H$ appears. It also means $v_m$ is connected to exactly two vertices $v_i,v_j, i,j<m$ and $v_i,v_j\in H$ . So, we have at least two Hamiltonian cycles in $H$ and both pass trough $v_m$ like $\dots v_i\, v_m\,v_j\,\dots$ . Since $v_i$ and $v_j$ are connected, by (ii) it yields there are also two Hamiltonian cycles in $H-v_m,$ contradiction.

Remark. More analysis and comments on this problem can be found on my blog.

This post has been edited 1 time. Last edited by dgrozev, Jul 27, 2022, 8:28 AM

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marcin

8 posts

marcin

#29 h Jul 28, 2022, 2:57 PM • 3 Y

Y by gvole, Mango247, Mango247

This is indeed a very nice problem. With not much more effort than what has been observed in previous posts one can show the following.

Theorem. Let $p_1<p_2$ be odd primes such that $p_1p_2=w^2+w+k$ for some $w\geq 0$ (such $w$ is unique, if it exists). Let $p$ be an odd prime. Then $pp_1$ and $pp_2$ are both of the form $x^2+x+k$ for some integer $x$ if and only if $p=p_1+p_2\pm (2w+1)$ ( $x=w\pm p_i$ works for $p$ and $p_i$ ).

This result can be used to construct sets $S$ of primes which can be arranged as in the problem. For example, starting with $w=1$ , $k=53$ , $p_1=5$ , $p_2=11$ we get a circle of primes: $5,13,101,43,11,19,107,269,37$ (in that order).

This post has been edited 1 time. Last edited by marcin, Jul 28, 2022, 2:59 PM
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62861

3564 posts

62861

#30 h Sep 5, 2022, 5:09 AM • 18 Y

Y by samrocksnature, rama1728, L567, a1267ab, megarnie, Justpassingby, mira74, pog, Quidditch, Grizzy, Miquel-point, CyclicISLscelesTrapezoid, PRMOisTheHardestExam, Assassino9931, EpicBird08, math_comb01, NO_SQUARES, MS_asdfgzxcvb

I don't really use this site anymore for various reasons, but I hope you all enjoyed the problem! Here's a postmortem, I suppose.

As you might guess, my only real regret with this problem is not noticing the glitch that arises with the statement having "positive integer $x$ " in it. That was not intentional, and I feel quite bad for the people who had to agonize over it. (Though apparently neither the PSC nor the jury noticed it either, so I don't feel that bad about not realizing that subtlety.)

Other than that, I'm really happy with the problem, and I'm glad that in the end, this is the problem of mine that made it onto the IMO. To be honest, I had my hopes: we all know how neglected N is at the IMO, and this problem is hard in a non-technical way: an enormous plus for gaining jury approval. But I didn't want to hope, since getting problems onto the IMO is basically a crapshoot.

Those of you who know me will know that I'm a big fan of "surprising" problems; I typically use USA TSTST 2019/9 as my main example. A lot of people don't seem to like this kind of problem (maybe because the actual math content is usually not inherently as interesting), but as a designer, I love providing that kind of experience, as well as receiving it. So it shouldn't come as a surprise that I really like this problem, too. Even after all these months, I still can't believe the main claim is true... it just feels so absurd, ya know? I'm really glad I was able to package the main claim into a nice extraction: if I just asked to prove the main claim, it would still be surprising, but much more boring.

I'm a bit surprised I haven't seen the main claim before, actually. It feels like the sort of thing that'd be famous in algebraic number theory circles, for example. But I guess not.

Fun fact I: the main claim is actually true for USAMO 2020/4 as well. To elaborate, spoilers!

Fun fact II: the first draft of this problem was the same, but it involved $x^2 + k$ instead of $x^2 + x + k$ . The solution is essentially the same, but after thinking about it for a bit, I realized that the problem was basically unusable unless I changed the expression $x^2 + k$ . Why?

Fun fact III: This statement might seem really stupid, in the sense that surely large prime rings don't exist at all. That's what I thought too, until I decided to see for myself. In the end, I ended up discovering a ring of 385 primes. I spent relatively little computer time discovering it (and had some inefficiencies in my scripting), so I invite everyone to try and beat that record! The 385-prime ring I found

With $k = 41$ :

53, 4357, 104173, 65921, 36383, 99527, 193789, 2089123, 1010357, 2465263, 319169, 15559, 3449, 2647, 1951, 152297, 542189, 119773, 91151, 66431, 222137, 1336799, 469069, 45613, 1047941, 656291, 355867, 146669, 874879, 2213327, 305119, 3336209, 1623467, 520963, 794201, 1124833, 28697, 15683, 42557, 6571, 39607, 1238833, 835421, 2653681, 5494387, 9357539, 511223, 1515317, 8868173, 114079681, 59334071, 22324807, 3051889, 5120939, 7722467, 266239, 693809, 3931783, 1322317, 100469, 13913, 74419, 23977, 1361, 62983, 935021, 512657, 1394849, 216259, 45827, 31393, 100787, 1193989, 600979, 209543, 357661, 545141, 19681, 10691, 28867, 165089, 2118023, 6271891, 12626693, 21182429, 1100467, 413089, 772867, 1244423, 1827757, 55889, 1558873, 5110711, 1024427, 601759, 290869, 91757, 951109, 452033, 136471, 190031, 4423, 9239, 15809, 24133, 115811, 275911, 34211, 877, 6653, 88001, 46261, 317741, 121523, 232439, 379009, 17827, 2699, 15937, 497729, 335539, 205223, 106781, 1394413, 4140947, 8346383, 43984757, 14010721, 21133961, 729451, 4997297, 1908223, 278051, 529747, 40213, 768107, 456821, 1325351, 225961, 1501921, 562763, 75527, 5519, 9337, 14153, 499, 1399, 2753, 14401, 94583, 245107, 35171, 397093, 195907, 2505623, 34680911, 18542791, 7415917, 144797293, 455529251, 86675291, 252704911, 43385123, 109207907, 204884269, 330414209, 14926789, 1300289, 486769, 2723989, 907757, 1458871, 65063, 4561, 124427, 81343, 252887, 2980139, 1496779, 3779057, 519193, 47381, 135283, 268267, 446333, 669481, 22541, 54167, 99439, 158357, 6823, 32497, 1390709, 998029, 670343, 5180017, 13936673, 2123491, 4391941, 407651, 209953, 77249, 867653, 427117, 141079, 9539, 227, 1439, 18679, 9749, 25453, 3697, 42139, 122327, 712303, 244261, 20873, 52051, 589997, 4310569, 1711069, 291563, 3731527, 11045429, 129098443, 64620427, 162661963, 22233269, 37295047, 1936969, 5033449, 725537, 1353973, 6964457, 2176871, 97231, 7001, 11351, 55673, 16747, 169003, 1218571, 479957, 2779783, 949609, 4975787, 1577959, 2365007, 3310753, 79349, 23189, 107209, 688907, 252583, 30677, 523, 941, 25981, 205103, 85087, 1011233, 509659, 178259, 950479, 6262847, 2333693, 305497, 3199319, 9148267, 1527563, 466801, 17033, 9967, 323003, 4724099, 14278309, 2576557, 1075021, 6462593, 2266021, 63922471, 209814503, 42117791, 131659867, 270892249, 24845153, 12104557, 3896003, 219491, 135913, 406397, 72269, 191689, 2197697, 1091273, 2727311, 368227, 1911661, 601883, 892657, 28559, 4783, 60497, 31259, 80909, 457697, 153733, 11587, 1481, 26161, 15193, 7187, 2143, 21517, 10079, 207643, 1604381, 657661, 126227, 372313, 2176331, 748337, 64969, 844867, 2507291, 29317943, 14677801, 36952793, 69332267, 111816223, 5052241, 8479717, 441263, 3020431, 1152751, 13179611, 38280013, 6536771, 16319657, 91442699, 30501409, 49082027, 72061511, 2199433, 167597, 317963, 23869, 2927, 3833, 17327, 110879, 285517, 40543, 4861, 21683, 50527, 565319, 277829, 687917, 3846023, 25542677, 174261149, 66370753, 9565711, 1280791, 91393, 6011, 7283, 31859, 8677, 10193, 43987, 11831, 13591, 127843, 358229, 58067, 15473, 65839, 17477, 74099, 19603, 82847, 21851, 61

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rama1728

800 posts

rama1728

#31 h Sep 5, 2022, 5:55 AM

Y by

CantonMathGuy wrote:

With $k = 41$ :

53, 4357, 104173, 65921, 36383, 99527, 193789, 2089123, 1010357, 2465263, 319169, 15559, 3449, 2647, 1951, 152297, 542189, 119773, 91151, 66431, 222137, 1336799, 469069, 45613, 1047941, 656291, 355867, 146669, 874879, 2213327, 305119, 3336209, 1623467, 520963, 794201, 1124833, 28697, 15683, 42557, 6571, 39607, 1238833, 835421, 2653681, 5494387, 9357539, 511223, 1515317, 8868173, 114079681, 59334071, 22324807, 3051889, 5120939, 7722467, 266239, 693809, 3931783, 1322317, 100469, 13913, 74419, 23977, 1361, 62983, 935021, 512657, 1394849, 216259, 45827, 31393, 100787, 1193989, 600979, 209543, 357661, 545141, 19681, 10691, 28867, 165089, 2118023, 6271891, 12626693, 21182429, 1100467, 413089, 772867, 1244423, 1827757, 55889, 1558873, 5110711, 1024427, 601759, 290869, 91757, 951109, 452033, 136471, 190031, 4423, 9239, 15809, 24133, 115811, 275911, 34211, 877, 6653, 88001, 46261, 317741, 121523, 232439, 379009, 17827, 2699, 15937, 497729, 335539, 205223, 106781, 1394413, 4140947, 8346383, 43984757, 14010721, 21133961, 729451, 4997297, 1908223, 278051, 529747, 40213, 768107, 456821, 1325351, 225961, 1501921, 562763, 75527, 5519, 9337, 14153, 499, 1399, 2753, 14401, 94583, 245107, 35171, 397093, 195907, 2505623, 34680911, 18542791, 7415917, 144797293, 455529251, 86675291, 252704911, 43385123, 109207907, 204884269, 330414209, 14926789, 1300289, 486769, 2723989, 907757, 1458871, 65063, 4561, 124427, 81343, 252887, 2980139, 1496779, 3779057, 519193, 47381, 135283, 268267, 446333, 669481, 22541, 54167, 99439, 158357, 6823, 32497, 1390709, 998029, 670343, 5180017, 13936673, 2123491, 4391941, 407651, 209953, 77249, 867653, 427117, 141079, 9539, 227, 1439, 18679, 9749, 25453, 3697, 42139, 122327, 712303, 244261, 20873, 52051, 589997, 4310569, 1711069, 291563, 3731527, 11045429, 129098443, 64620427, 162661963, 22233269, 37295047, 1936969, 5033449, 725537, 1353973, 6964457, 2176871, 97231, 7001, 11351, 55673, 16747, 169003, 1218571, 479957, 2779783, 949609, 4975787, 1577959, 2365007, 3310753, 79349, 23189, 107209, 688907, 252583, 30677, 523, 941, 25981, 205103, 85087, 1011233, 509659, 178259, 950479, 6262847, 2333693, 305497, 3199319, 9148267, 1527563, 466801, 17033, 9967, 323003, 4724099, 14278309, 2576557, 1075021, 6462593, 2266021, 63922471, 209814503, 42117791, 131659867, 270892249, 24845153, 12104557, 3896003, 219491, 135913, 406397, 72269, 191689, 2197697, 1091273, 2727311, 368227, 1911661, 601883, 892657, 28559, 4783, 60497, 31259, 80909, 457697, 153733, 11587, 1481, 26161, 15193, 7187, 2143, 21517, 10079, 207643, 1604381, 657661, 126227, 372313, 2176331, 748337, 64969, 844867, 2507291, 29317943, 14677801, 36952793, 69332267, 111816223, 5052241, 8479717, 441263, 3020431, 1152751, 13179611, 38280013, 6536771, 16319657, 91442699, 30501409, 49082027, 72061511, 2199433, 167597, 317963, 23869, 2927, 3833, 17327, 110879, 285517, 40543, 4861, 21683, 50527, 565319, 277829, 687917, 3846023, 25542677, 174261149, 66370753, 9565711, 1280791, 91393, 6011, 7283, 31859, 8677, 10193, 43987, 11831, 13591, 127843, 358229, 58067, 15473, 65839, 17477, 74099, 19603, 82847, 21851, 61

Nice! Why don't you use AoPS though...

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little-fermat

147 posts

little-fermat

#32 h Sep 12, 2022, 3:04 PM • 1 Y

Y by Mango247

I have discussed this problem on my YouTube channel "little fermat".
Video link: Video Solution

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Ramanujan1057894736842

15 posts

Ramanujan1057894736842

#33 h Mar 10, 2023, 2:17 AM

Y by

Can someone please verify this solution
Let $k$ be a positive integer and let $S$ be a finite set of odd prime numbers. Prove that there is at most one way (up to rotation and reflection) to place the elements of $S$ around the circle such that the product of any two neighbors is of the form $x^2+x+k$ for some positive integer $x$ .

Let $p_1,p_2,\ldots,p_n$ be the odd prime numbers in $S$ , listed in clockwise order around the circle. For each $i=1,2,\ldots,n$ , let $x_i$ be a positive integer such that $p_{i-1}p_{i+1}=x_i^2+x_ik$ (where we interpret $p_0=p_n$ and $p_{n+1}=p_1$ ). We want to show that the values $x_1,x_2,\ldots,x_n$ uniquely determine the placement of the primes around the circle, up to rotation and reflection.

First, note that $x_i$ is uniquely determined by $p_{i-1}$ and $p_{i+1}$ . Indeed, if $x_i'$ is another positive integer such that $p_{i-1}p_{i+1}=x_i'^2+x_i'k$ , then $x_i^2+x_ik=x_i'^2+x_i'k$ , so $x_i=x_i'$ .

Let $S'$ be the set of positive integers $x_1,x_2,\ldots,x_n$ . Note that if we have two different circle arrangements of $S$ that satisfy the given condition, then the corresponding sequences of $x_i$ 's must be the same (up to cyclic permutation). Otherwise, there would be some $i$ such that $x_i$ in one sequence differs from $x_i$ in the other sequence, and we could use this to find two neighboring primes whose product is not of the form $x^2+x+k$ .

Therefore, it suffices to show that the sequence $S'$ uniquely determines the circle arrangement of $S$ up to rotation and reflection. We will prove this by induction on $n$ . The base case $n=1$ is trivial.

Now suppose $n>1$ and the claim is true for smaller values of $n$ . Without loss of generality, we may assume that $p_1=3$ . Then $p_2$ is odd and we have $p_1p_2=x_1^2+x_1k$ , so $x_1$ is uniquely determined by $p_2$ . Note that $x_1>1$ , since $p_1p_2=x_1(x_1+k)\geq 2x_1$ . Therefore, we can define a new sequence $S''$ by replacing the first element $x_1$ of $S'$ with $x_1-1$ . By induction, the sequence $S''$ determines the circle arrangement of the remaining primes (listed in clockwise order from $p_2$ ). Moreover, we can recover $x_1$ from $S''$ and $p_2$ using the equation $p_1p_2=(x_1-1+k)(x_1-1)+x_1-1$ , so the circle arrangement of $p_1$ and $p_2$ is also determined.

Therefore, the sequence $S'$ determines the circle arrangement of $S$ up to rotation and reflection, as desired.

This post has been edited 1 time. Last edited by Ramanujan1057894736842, Mar 10, 2023, 2:20 AM
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peppapig_

280 posts

peppapig_

#34 h May 19, 2023, 3:18 PM • 5 Y

Y by player01, john0512, Significant, mulberrykid, EpicBird08

Claim 1.
Fix set $S$ and the integer $k$ and let $q$ be the largest prime in $S$ . We claim that if there is such a circular configuration (as described in the problem), then the numbers next to $q$ are fixed.

1. We claim that there are at most $2$ integers $0\leq{}x\leq{}q$ such that $x^2+x+k\equiv{}0$ (mod $q$ ). If there is no such number, then we are done. Say there is such a number that we know works (where "works" means that $x^2+x+k\equiv{}0$ (mod $q$ )), call it $a$ . If $a$ is the only number that works. Otherwise, let there be another number, $b$ such that $b$ "works". We claim that $b$ is either $a$ or $q-1-a$ mod $q$ .

We have that $a^2+a+k\equiv{}b^2+b+k$ (mod $q$ ). Subtracting, we get
$(a^2-b^2)+(a-b)\equiv{}0 \text{ (mod } q)\iff{}(a-b)(a+b+1)\equiv{}0 \text{ (mod } q)$ and since $q$ is prime, this implies that either $q\mid{}a-b$ or $q\mid{}a+b+1$ , meaning that $b$ is either $a$ or $q-1-a$ mod $q$ .

2. Let the two primes next to $q$ be $p$ and $r$ . We claim that if we let $pq=a^2+a+k$ and $qr=b^2+b+k$ , we must have $a+b=q-1$ . Notice that if $p,r<q$ , we have that $a,b<q$ which immediately rules out the possibility that $a\equiv{}b$ mod $q$ . Since $a+b\leq{}2q-2$ , we also have that $a+b=q-1$ , since the sum cannot be $2q-1$ or greater, and the sum must be $-1$ mod $q$ .

3. Since by $\textcircled{1}$ and $\textcircled{2}$ , $a$ and $b$ are unique if they exist, we have that $p$ and $r$ must also be unique, proving our claim.

Claim 2.
We claim that if we have three primes such that $p,r<q$ and $pq$ and $qr$ are both expressible as $x^2+x+k$ for some positive integer $x$ , then $pr$ is also expressible as $x^2+x+k$ for some $x$ .

1. We will prove the claim that "if $x^2+x+k=pq$ and $(q-1-x)^2+(q-1-x)+k=qr$ **, then $\frac{(q-1-x)^2+(q-1-x)+k}{(x-p)^2+(x-p)+k}=\frac{q}{p}$ . Note that if this claim is true, it immediately follows that $(x-p)^2+(x-p)+k=pr$ , finishing the proof of Claim 2.

**Recall that from Claim 1, if $f(x)=x^2+x+k=pq$ , then $f(q-1-x)=qr$ , therefore $(q-1-x)^2+(q-1-x)+k=qr$ .

2. Expanding the above equation and replacing all $x^2+x+k$ with $pq$ , we end up with $\frac{x^2+x+k+p^2-2px-p}{x^2+x+k+q^2-2qx-q}=\frac{pq+q^2-q(2x+1)}{pq+p^2-p(2x+1)}=\frac{q(p+q-2x-1)}{p(p+q-2x-1)}$ which is clearly equal to $\frac{q}{p}$ , proving our claim.

Inductive Step.
First, notice that if there are $2$ primes in $S$ , then there is trivially at most $1$ way, and we would be done.

Consider any set with more than 3 elements. Suppose S has a configuration satisfying the problem conditions. Now take out the largest prime number from that configuration. By combining Claims 1 & 2, then remaining prime numbers must also be a circular configuration that satisfies the problem conditions. Since if there is a circular configuration that satisfies the problem conditions, the largest prime must have a fixed spot that it can be placed in in the circle (if there is a valid one), meaning the new set $S$ with the largest prime has at most the number of configurations the set without the largest prime did.

Since a set $S$ with two primes has at most $1$ configuration, any set $S$ has at most $1$ configuration, and we are done.

This post has been edited 6 times. Last edited by peppapig_, May 21, 2023, 5:16 PM
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Leo.Euler

577 posts

Leo.Euler

#37 h Jul 15, 2023, 10:55 PM • 3 Y

Y by Billybillybobjoejr., trk08, OronSH

Now this is the best ISL number theory problem I have done as of now. It possesses some really nice ideas and motivations behind them. Here we prove the result where the condition holds for nonnegative integers $x$ . In what follows, we prove two number theoretic claims and then induct on the logical graph formulation whence we apply those claims.

We say that a pair of primes $(p, q) \in S^2$ is representative if there exists an nonnegative integer $x$ such that $pq=x^2+x+k$ .

Claim: Given a prime $p \in S$ , there are at most two distinct primes smaller than $p$ in $S$ such that they form representative pairs with $p$ .
Proof. Suppose that $(p, q)$ and $(p, r)$ are representative pairs; it suffices to show that no prime $s$ different from $p, q, r$ less than $p$ is in $S$ such that $(p, s)$ is representative. We can write $pq=x^2+x+k$ $pr = y^2+y+k$ for nonnegative integers $x$ and $y$ . Subtracting the equations, we have $p(q-r)=(x-y)(x+y+1)$ . Since $x$ and $y$ have to both be less than $p$ (as otherwise the RHS of both of the above equalities will clearly exceed $p^2$ ), we must have $-p<x-y<p$ , and $x-y \neq 0$ as $q-r \neq 0$ . Thus, $p \nmid x-y$ , so $p \mid x+y+1$ . But since $0 \le x, y < p$ , we must have $x+y+1=p.$ Now since $(p, s)$ is representative, there is a nonnegative integer $z$ such that $ps=z^2+z+k$ for some nonnegative integer $z$ . We analogously have $x+z+1=p, y+z+1=p,$ so $x=y=z$ , but this is a contradiction as $q, r, s$ are distinct. :yoda:

Claim: Given a prime $p \in S$ and distinct primes $q, r \in S$ smaller than $p$ such that $(p, q)$ and $(p, r)$ are representative, $(q, r)$ is also representative.
Proof. Let $pq=x^2+x+k$ and $pr=y^2+y+k$ for nonnegative integers $x$ and $y$ . By subtracting the two equations and applying appropriate bounding on $x, y$ as in the previous proof, we obtain $p=x+y+1$ . Now let $a=x+\tfrac{1}{2}$ , $b=y+\tfrac{1}{2}$ , and $\ell = k - \tfrac{1}{4}$ . Then $a+b=p$ , and $pq=a^2+ \ell,$ $pr = b^2 + \ell.$ By the Brahmagupta-Fibonacci identity, it follows that $p^2qr = (pq)(pr) = (ab-\ell)^2 + (a+b)^2\ell = (ab - \ell)^2+\ell p^2,$ so $qr = (pq)(pr)/(p^2) = \left(\frac{ab-\ell}{p}\right)^2 + \ell.$ I contend that $\tfrac{ab-\ell}{p}-\tfrac{1}{2}$ is an integer.

Note that $-2xy \equiv (x+y)(x+y+1)-2xy \equiv (x^2+x)+(y^2+y) \equiv -2k \pmod{p}$ , so $xy \equiv k \pmod{p}$ . Thus, $p \bigg(\frac{ab-\ell}{p}-\frac{1}{2}\bigg) = ab-\ell - \frac{p}{2} = (x+1/2)(y+1/2) - k + 1/4 - \frac{p}{2} = xy - k \equiv 0 \pmod{p},$ so $\tfrac{ab-\ell}{p}-\tfrac{1}{2}$ is an integer, as claimed. This means that $(q, r)$ is representative, as desired. :yoda:

Now we take the graph formulation of the problem. Let the elements of $S$ have corresponding vertices $v_1, \ldots, v_n$ in order of size. Then consider a graph $G$ on those vertices such that, (1) for each $1< i \le n$ , at most two of the vertices among $v_1, \ldots, v_{i-1}$ are connected with $v_i$ , and (2) for each $1< i \le n$ , if $v_i$ is connected to two vertices $v_j, v_k$ where $j \neq k$ and $j, k < i$ , then $v_j$ is connected to $v_k$ . From the previous two claims, the problem can be reformulated as the following lemma.

Lemma: Any subgraph $H$ of $G$ has at most one Hamiltonian cycle.
Proof. Induct on $n$ , where the base cases $n=1, 2, 3$ are trivially true. Consider $G$ with $n=k$ , and suppose the lemma holds for this graph. Then add a vertex $v_{k+1}$ to the graph, and connect to it at most two vertices, and in the case of $v_{k+1}$ having degree $2$ , connect its two neighbors. By inductive hypothesis, we only have to consider the Hamiltonian cycles of subgraphs containing $v_{k+1}$ . If $\deg(v_{k+1}) \le 1$ , then all such subgraphs do not have a Hamiltonian cycle. On the other hand, if $\deg(v_{k+1})=2$ , and if we label the neighbors of $v_{k+1}$ as $u$ and $v$ , then it suffices to show that $u$ and $v$ are not contained in a subgraph with more than $1$ Hamiltonian cycle; however, this is clear by inductive hypothesis. This completes the induction. :yoda:

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IAmTheHazard

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IAmTheHazard

#38 h Sep 30, 2023, 10:05 PM

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what the heck z antiproblem. modern day miracle.

Relax the condition to $x$ being a nonnegative integer. We induct on $|S|$ (with $k$ fixed), with $|S| \leq 3$ being essentially vacuous. Suppose $|S| \geq 4$ and the statement has been proven for $|S|-1$ .

Suppose that an arrangement exists. Let $p$ be the largest prime in $S$ . Then there are at most two solutions modulo $p$ to $x^2+x+k \equiv 0 \pmod{p}$ , hence at most two solutions $0 \leq x_2<x_1 \leq p-1$ . We have the following.

Claim 1: In any arrangement, if $q$ is adjacent to $p$ , then we must have $pq=x_i^2+x_i+k$ for some $i \in \{1,2\}$ .
Proof: Suppose $pq=r^2+r+k$ where $r \geq 0$ . Then $r \equiv x_1,x_2 \pmod{p}$ . On the other hand, if $r \geq p$ , then $r^2+r+k>p^2>pq$ : contradiction. $\blacksquare$

Since we have at most $2$ choices for $i$ , the two primes $q_1>q_2$ adjacent to $p$ are fixed in any configuration, and we have $pq_i=x_i^2+x_i+k$ for $i=1,2$ . We now prove the following key claim.

Claim 2: In fact, $q_1q_2$ is of the form $x^2+x+k$ for some nonnegative $x$ as well.
Proof: First subtract the two given equations to obtain $p(q_1-q_2)=(x_1-x_2)(x_1+x_2+1)$ . $p$ must divide some term of the RHS, but it can't be $x_1-x_2 \leq p-1$ , hence $p \mid x_1+x_2+1$ . On the other hand, since $x_1+x_2+1 \leq 2p-1$ , it follows that $x_1+x_2+1=p$ , so $x_1-x_2=q_1-q_2$ . Solving for $x$ yields $x=\tfrac{(p+q_1+q_2)-1}{2}$ , hence
$\frac{(p+q_1+q_2)^2-1}{4}+k=pq_1 \implies -k=\frac{(p^2+q_1^2+q_2^2)-2q_1q_2-2p(q_1+q_2)-1}{4}.$ We now consider the determinant of $x^2+x-(q_1q_2-k)$ ; it equals
$1+4(q_1q_2-k)=1+4\left(q_1q_2+\frac{(p^2+q_1^2+q_2^2)-2q_1q_2-2p(q_1+q_2)-1}{4}\right)=1+(p^2+q_1^2+q_2^2)+2q_1q_2-2p(q_1+q_2)-1=(p-(q_1+q_2))^2.$ Since $p,q_1,q_2$ are odd, this is odd as well, and since it's a square it follows that $x^2+x+k=q_1q_2$ has a nonnegative integer solution by the quadratic formula. $\blacksquare$

Then if we have two different arrangements of the elements of $S$ (both which have $q_1,q_2$ adjacent to $p$ ), then delete $p$ in both to get two different arrangements of the elements of $S \setminus \{p\}$ , contradicting the inductive hypothesis. $\blacksquare$

This post has been edited 2 times. Last edited by IAmTheHazard, Sep 30, 2023, 10:11 PM

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Inconsistent

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Inconsistent

#39 h Nov 28, 2023, 1:37 AM

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Beautiful!

Downward induct in triangles: key claim is that the two smaller children of any new integer (adding prime numbers from least to greatest) must themselves be children of one another. (Where children means an edge to a smaller prime and two primes are connected by an edge if $pq - (4k-1)$ is a perfect square. The result can be obtained by a clever Vieta-esque jump on $x^2 - 2(a+b)x + (a-b)^2 + (k - 1) = 0$ or outright algebraic bash. Truly quite elegant! To end, simply eliminate the largest element of the cycle and claim uniqueness.

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awesomeming327.

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awesomeming327.

#40 h Dec 18, 2023, 5:36 PM

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First, we'll reduce the condition on $x$ to nonnegative integers. Call an unordered pair $(p,q)$ of distinct odd prime numbers good if $pq$ is of the form $x^2+x+k$ for some nonnegative integer $x$ . We claim that if $(p,q)$ and $(p,r)$ are good, and $p>q,r$ then $(q,r)$ is good. Let $pq=a^2+a+k$ , $pr=b^2+b+k$ . Then, $p(q-r)=(a-b)(a+b+1)$ . If $p\mid a-b$ then $a+b+1\mid q-r$ so $p\le a-b<a+b+1\le q-r<p$ , a contradiction. Thus, $p\mid a+b+1$ . Let $a+b+1=ps$ then
$4p^2=2p(q+r)=2(a^2+b^2+a+b+2k)>(a+b)^2+2a+2b+4k=(a+b)(a+b+2)+4k>(a+b+1)^2=p^2s^2$ so $s^2<4$ implying that $s=1$ . Thus, $a+b+1=p$ , so $q-r=a-b$ . Let $a=q+t$ and $b=r+t$ then $p=q+r+2t+1$ and $q(q+r+2t+1)=(q+t)^2+(q+t)+k$ which implies $qr=t^2+t+k$ , as desired.

Suppose that the smallest set for which there are at least two different ways whose elements can be placed in a way that satisfies the conditions mentioned is $S$ , then in each situation, we can remove the largest prime. By our claim, the remaining circle still satisfies the conditions. Since we remove largest prime, there are at most two solutions to $x^2+x+k$ being divisible by that largest prime, so the neighbors of the prime we removed are known. However, by minimality we must be left with the same configuration, contradiction.

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Acclab

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Acclab

#41 h Jul 13, 2024, 8:45 PM

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For each $p_1$ with neighbors $p_2$ and $p_3$ , such $k$ exists so both $p_1p_2 = a^2+a+k$ and $p_1p_3 = b^2+b+k$ for some positive integer $a, b$ (where WLOG $p_3>p_2$ ) if and only if
$p_1(p_3-p_2) = (b-a)(b+a+1).$ for some $a$ , $b$ . If furthermore $p_1>p_3>p_2$ , then a size argument is available. Since $p_1 | b-a$ implies $b+a+1 \leq p_3-p_2 < p_1 \leq b-a$ , a contradiction, we know $b+a+1 = rp_1$ and $b-a = \frac{p_1}{r}$ for some $r \in \mathbb{N}$ . Solving we get:
$a = \frac{rp_1-\frac{p_3-p_2}{r}-1}{2}, b = \frac{rp_1+\frac{p_3-p_2}{r}-1}{2}.$ This implies $b \geq p_1$ if $r \geq 2$ , which yields $b^2+b+k > p_1^2 > p_1p_3$ , a contradiction. Thus
$a = \frac{p_1-p_3+p_2-1}{2}, b = \frac{p_1+p_3-p_2-1}{2}.$ Calculation gives $k = \frac{2p_1p_2+2p_2p_3+2p_1p_3 - p_1^2 - p_2^2 - p_3^2 + 1}{4}$ . That is to say, $k$ is uniquely determined by the values of $p_1, p_3, p_2$ .

Notice that $k$ is symmetric to $p_1, p_2, p_3$ , and without the size argument, the solutions to $a, b, k$ also works. This idea gives rise to the following claim.

Claim. Fix $k \in \mathbb{N}$ . Denote $p_1 \sim p_2$ if $p_1p_2 = x^2+x+k$ for some $x$ . Then if $p_1>p_3>p_2$ and both $p_1 \sim p_2$ and $p_1 \sim p_3$ , then we also have $p_2 \sim p_3$ .

Proof. Let $c = \frac{p_2+p_3-p_1-1}{2}.$ Since $a^2+a+k = p_1p_2$ and $b^2+b+k = p_1p_3$ , we have that $c^2+c+k = p_2p_3$ if and only if the sum of the three formulae holds: $(\frac{p_1+p_2-p_3}{2})^2 - \frac{1}{4} + k + (\frac{p_1-p_2+p_3}{2})^2 - \frac{1}{4} + k + (\frac{-p_1+p_2+p_3}{2})^2 - \frac{1}{4} + k = p_1p_2 + p_2p_3 + p_1p_3$ by plugging in the values of $a,b$ derived via the size argument. That reduces to $k = \frac{2p_1p_2+2p_2p_3+2p_1p_3 - p_1^2 - p_2^2 - p_3^2 + 1}{4}$ , agreeing with the unique solution to $k$ given that $p_1 \sim p_2$ and $p_1 \sim p_3$ , proving such choice of $c$ works, and thus $p_2 \sim p_3$ if $c>0$ .

If $c<0$ , we instead let $c \mapsto -1-c$ so $c^2+c$ agrees by Vieta. ( $p_1=p_2+p_3$ is impossible for all these primes are odd. )

Remark. Denote $a \sim^kb$ to specify the value of $k$ . A more heuristic way of discovering this claim is by noticing that solving $p_1 \sim^k p_3$ , $p_1 \sim^k p_2$ gives (uniquely) $a = \frac{p_1-p_3+p_2-1}{2}, b = \frac{p_1+p_3-p_2-1}{2}$ and $k = \frac{2p_1p_2+2p_2p_3+2p_1p_3 - p_1^2 - p_2^2 - p_3^2 + 1}{4}$ with the size condition. Notice $k$ is symmetric in terms of $p_1, p_2, p_3$ , and this is still a valid solution without the size condition, so shuffling the orders of $p_1, p_2, p_3$ would result in $c = \frac{p_2+p_3-p_1-1}{2}$ with the same $k$ as one solution to, say, $p_1 \sim^k p_2$ and $p_2 \sim^k p_3$ . Since the value of $k$ agrees, this is exactly what we have in the claim.

The rest follows by induction on $|S|$ . $|S| = 3$ is trivial. For $|S| \geq 4$ , suppose $S$ admits the largest prime $p_1$ . If in both configuration $p_1$ are adjacent to the same $p_2, p_3$ , then $p_2 \sim p_3$ by our claim and thus we can remove $p_1$ from the set and replace the two configuration by removing $p_1$ from the circles while keeping their validity. If instead $p_1$ are adjacent to at least three different neighbors, say $p_2, p_3, p_4$ . Denote $p_1p_2 = x^2+x+k$ , then since $p_1\sim p_3$ and $p_1 \sim p_4$ , it follows $x = \frac{p_1-p_3+p_2-1}{2} = \frac{p_1-p_4+p_2-1}{2}.$ Since these two values of $x$ admits different $x^2+x+k$ , we have a contradiction.

This post has been edited 3 times. Last edited by Acclab, Jul 13, 2024, 9:07 PM

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Mr.Sharkman

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Mr.Sharkman

#42 h Nov 8, 2024, 1:48 PM

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We will prove this through induction. For $n=2,$ notice that there is only $1$ way to arrange the primes around the circle regardless of the condition, so this case is obvious.

Assume that this is true for $n.$ Now, take a set of $n+1$ primes in a configuration that works. Assume that $p$ is the largest prime, and let $q$ and $r$ be primes next to it.

Claim: If we delete $p$ from the circle, we still have a valid configuration.
Proof: Let $x^{2}+x+k = pq,$ and $y^{2}+y+k = pr.$
Then, $x^{2} < x^{2}+x+k =pq \le p^{2},$ so $x < p.$ Similarly, $y < p.$ Now, $4(x^{2}+x+k) = (2x+1)^{2}+(4k-1),$ so $x,y$ are the two unique solutions less than $p$ such that $(2x+1)^{2} \equiv (1-4k) \pmod p.$ This means that $(2x+1)+(2y+1)=2p.$ Now, let $m = 4k-1,$ and let $a = 2x-1.$ Then,
$qr = \frac{pr \cdot pq}{p^{2}} = \frac{(a^{2}+m)((2p-a)^{2}+m)}{16p^{2}}.$ So,
$qr = \frac{a^{2}(2p-a)^{2}+ma^{2}+m(2p-a)^{2}+m^{2}}{16p^{2}}.$ Thus,
$qr = \frac{m(2a^{2}-4ap+4p^{2})+m^{2}+a(2p-a)}{16p^{2}} = \frac{1}{4}\left(\frac{a(2p-a)-pm}{p}\right)^{2}+\frac{m}{4}$ so $qr$ is of the form $x^{2}+x+k,$ as desired. Thus, we have proven the claim.

Now, notice that, if we add $p$ back to our set of primes, it must be next to $q$ and $r$ (as there are only $2$ values of $x,y$ which have $p \mid x^{2}+x+k, y^{2}+y+k,$ and $q$ and $r$ correspond to $x,y$ ). Thus, we have finished the inductive step.

This post has been edited 2 times. Last edited by Mr.Sharkman, Nov 8, 2024, 1:52 PM

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bin_sherlo

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bin_sherlo

#43 h Mar 9, 2025, 11:45 AM • 1 Y

Y by farhad.fritl

We will show that the statement holds for nonnegative integers. Let $|S|=n$ . We will induct on $n$ . First, note that it holds for $n\leq 3$ . Suppose that it's true for $n-1$ . Let $p$ be the largest prime in $S$ . If $p_ip_j=x^2+x+k$ for some nonnegative integer, then call $p_i,p_j$ friends.

Claim: $p$ has at most two friends.
Proof: Suppose that $pq=x^2+x+k, \ pr=y^2+y+k, \ ps=z^2+z+k$ for $p>q>r>s$ . We have $p(q-r)=(x-y)(x+y+1)$ and if $p|x-y$ , then $x+y+1|q-r$ which is impossible since $p\leq x-y<x+y+1<q-r<p$ does not hold. Thus, $p|x+y+1$ . Similarily $p|y+z+1,z+x+1$ but these give that $p|(x+z+1)-(y+z+1)=x-y$ which contradicts.

Claim: If $p$ is friend with both $q>r$ , then $q$ and $r$ are also friends.
Proof: We have $p|x+y+1$ and $x-y|q-r$ . Set $pl=x+y+1$ and $(x-y)l=q-r$ . If $l>1$ , then
$\frac{(x+y+1)^2}{2}\geq 2p^2>p(q+r)=x^2+y^2+x+y+2k\geq x^2+y^2+x+y+2$ However, $x^2+y^2+1+2xy+2x+2y>2x^2+2y^2+2x+2y+4$ implies $(x-y)^2+3<0$ which is impossible. Hence $l=1$ . We see that $p=x+y+1$ and $x-y=q-r$ . Thus, $x=\frac{p+q-r-1}{2}$ and $y=\frac{p-q+r-1}{2}$ .
$pq=(\frac{p+q-r-1}{2})(\frac{p+q-r+1}{2})+k\iff p^2+q^2+r^2+4k-1=2pq+2qr+2rp$ $p^2+q^2+r^2+4k-1=2pq+2qr+2rp\iff qr=(\frac{p-q-r-1}{2})(\frac{p-q-r+1}{2})+k$ If $p+1\geq q+r$ , then $qr=a^2+a+k$ for $a=\frac{p-q-r+1}{2}$ and if $p+1\leq q+r$ , then $a=\frac{q+r-p-1}{2}$ .

Since there is at most one hamiltonian cycle in $S \ \setminus \{p\}$ and $p$ must be neighbours with $q,r$ in a hamiltonian cycle in $S$ , we get that there is at most one hamiltonian cycle in $S$ as desired. $\blacksquare$

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awesomehuman

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awesomehuman

#44 h Apr 28, 2025, 5:08 PM

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We will prove the statement with "positive" replaced by "nonnegative".

Let $n$ be the number of elements of $S$ . We will use induction on $S$ . The base case $n=3$ is obvious. Let $n>3$ .
Create a graph where $p$ is connected to $q$ if $pq$ is of the form $x^2+x+k$ . Assume toward a contradiction there is more than $1$ Hamiltonian cycle.

Let $p$ be the maximum element of $S$ . Let $T = S\setminus \{p\}$ .

Let $q, r<p$ be $p$ 's neighbors.

Let $pq = a^2+a+k$ and $pr = b^2+b+k$ . Then, $a, b \leq p$ . There are at most two roots of $x^2+x+k\equiv 0\pmod{p}$ , so $a$ and $b$ are those roots. So, $p$ must be connected to $q$ and $r$ in any Hamiltonian cycle.

Because $p-a-1$ is a root of $x^2+x+k\equiv 0\pmod{p}$ , $b = p-a-1$ . So, $a+b = p-1$ .
We have
$p(q-r) = (a-b)(a+b+1)$ $p(q-r) = (a-b)p$ $q-r = a-b$ $a = \frac{p+q-r-1}2$ $b = \frac{p-q+r-1}2.$ So,
$pq = \left(\frac{p+q-r-1}2\right)^2 + \left(\frac{p+q-r-1}2\right) + k$ $-4k = p^2+q^2+r^2-2pq-2qr-2pr$ $rq = \left(\frac{r+q-p-1}2\right)^2 + \left(\frac{r+q-p-1}2\right) + k.$ $rq = \left(\frac{p-r-q-1}2\right)^2 + \left(\frac{p-r-q-1}2\right) + k.$ One of $\frac{p-r-q-1}2$ and $\frac{r+q-p-1}2$ is nonnegative.
So, $r$ and $q$ are connected.
In any Hamiltonian cycle, $p$ is connected to $q$ and $r$ , so we can delete $p$ and connect $q$ and $r$ . This reduces to the $n-1$ case.

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