Iran second round 2025-q1

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mohsen

#1 h Apr 19, 2025, 10:21 AM • 1 Y

Y by sami1618

Find all positive integers n>2 such that sum of n and any of its prime divisors is a perfect square.

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#2 h Apr 19, 2025, 5:29 PM • 1 Y

Y by Sedro

Answer: No such number exists.

Solution. We start by proving the following two claims:

Claim 1. We have that $n$ is squarefree.
Proof. If there exists a prime $p$ for which $p^2|n$ , then $n+p$ must be a perfect square. But $v_p(n+p)=1$ , a contradiction.

Claim 2. We have that $n$ has at most two distinct prime divisors.
Proof. Assume otherwise that $p$ , $q$ , and $r$ are three distinct prime divisors of $n$ . Without loss of generality, assume that $p<q<r$ . It is easy to see that both $p$ and $q$ must be strictly smaller than $\sqrt{n}$ . On the other hand, $\sqrt{n+q}-\sqrt{n+p}\geq 1\Rightarrow q-p\geq \sqrt{n+q}+\sqrt{n+p}\geq 2\sqrt{n},$ which is a contradiction.

By the Claims, we must have that either $n=p$ for some prime $p>2$ or $n=pq$ for some distinct primes $p$ and $q$ . In the first case, $2p$ must be a perfect square, which is absurd. In the second case, both $p(q+1)$ and $q(p+1)$ must be perfect squares. This implies that $p|q+1$ and that $q|p+1$ . Assume that $p<q$ . Then it must be that $p=q-1$ which implies that $p=2$ and $q=3$ , which fails to work.

This post has been edited 1 time. Last edited by sami1618, Apr 19, 2025, 5:33 PM

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missionjoshi.65

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missionjoshi.65

#3 h Apr 20, 2025, 10:43 AM • 2 Y

Y by Calc_1, sami1618

Posting though very similar as above!
Claim 1: $n$ can't be a power of prime.

Suppose, $n = p^k$ for some prime $p$ , then we have $v_p(p^k+p) =1+v_p(p^{k-1}+1),$ which can't be a square when $k \geq 2$ . And, $k = 1$ gives $n = 2$ which is a contradiction.

Claim 2: $n$ cannot have three or more prime factors.

Suppose $p,q,r | n$ and $p<q<r$ , then

$n +p = a^2 \quad \text{and} \quad n+q = b^2$
This implies:
$q-p = (b-a)(b+a)$
Similarly,
$q-p \geq b+a = \sqrt{n+p} + \sqrt{n+q} > \sqrt{n}+\sqrt{n} = 2 \sqrt{n}$
which is a contradiction since both $p, q < \sqrt{n}.$

Now, for two primes $p<q$ we have $p|q+1$ and $q|p+1$ which implies $p=q-1,$ so only possible values are $p=2$ and $q=3$ , which do not work. And, we are done!

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#4 h Apr 22, 2025, 4:04 PM • 2 Y

Y by Mahdi_Mashayekhi, amirhsz

Here's a much different and overkill approach.
suppose $n$ has at least two different primes $q,r$ such that $q \equiv r \equiv 3 \mod 4$ . Note that $n$ is also square-free.
So we get
$q(\frac{n}{q}+1) = x^2 \Rightarrow (\frac{r}{q})=1$ Similarly $(\frac{q}{r})=1$ .
Now note that $(\frac{r}{q})(\frac{q}{r})=(-1)^{\frac{r-1}{2}. \frac{q-1}{2}}=-1 \Rightarrow 1=(\frac{r}{q})=-(\frac{q}{r})=-1$ which is a contradiction.
if $n = 2p$ , we must have $3p$ is a perfect square which gives $p=3$ which doesn't work.
Now $n$ odd won't work because of mod $4$ . so $n$ is even. Consider a prime divisor $p$ of $n$ such $p \equiv 1 \mod 4$ we get $p(\frac{n}{p}+1) =x^2$ but $\frac{n}{p}+1 \equiv 3 \mod 4$ .
Thus we conclude that no such $n$ exists.

This post has been edited 2 times. Last edited by Parsia--, May 10, 2025, 3:18 PM
Reason: Shortened the solution

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1555 posts

#5 h Apr 23, 2025, 12:31 AM

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Suppose FTSOC there existed such $n$ , then if $n$ was not squarefree then there exists a prime $p$ with $p^2 \mid n$ however we have $n+p$ is a perfect square and thus $2 \mid \nu_p(n+p)=1$ , a clear contradiction!, therefore $n$ is squarefree.
Now let $n=p_1 \cdot p_2 \cdots p_k$ where $p_1<p_2 \cdots p_k$ . Clearly $k \ge 2$ as if $k=1$ then $2p_1$ is a perfect square but $p_1 \ne 2$ in this case so this isn't true, and if $k \ge 3$ then notice we must have $\sqrt{n+p_2}-\sqrt{n+p_1} \ge 1$ which implies that $n+p_2 \ge n+p_1+2\sqrt{n+p_1}+1$ which implies that $p_2^2>(p_2-p_1-1)^2 \ge 4n+4p_1>4p_2^2+4p_1$ which is clearly a contradiction.
And now if $n=p_1p_2$ then $p_2(p_1+1), p_1(p_2+1)$ are both perfect squares then $p_2 \mid p_1+1$ and $p_1 \mid p_2+1$ so $p_1p_2 \mid p_1p_2+p_1+p_2+1$ which means $p_1p_2 \mid p_1+p_2+1$ which means that $2 \le (p_1-1)(p_2-1) \le 2$ where equality is attained when $p_1=2$ and $p_2=3$ however $2 \cdot 4=8$ is not a perfect square thus contradiction either way and we are done :cool:

:cool:

This post has been edited 1 time. Last edited by MathLuis, Apr 23, 2025, 12:31 AM

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#6 h Apr 23, 2025, 4:52 AM

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Here you can find some solutions of second round
https://t.me/matholampiad123

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#7 h Apr 28, 2025, 7:50 AM

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Answer:such number do not exist.
Suppose on the contrary that a number $n$ satisfying the properties exists, and let $p$ and $q$ be two prime divisors of $n$ . Then by the hypothesis there exist $x,y \in \mathbb{N}$ such that $n+p=x^2$ and $n+q=y^2$ . clearly we can see that $p|x$ and $p|y$ . Then by a size argument it follows:
$p-q=x^2-y^2=(x-y)(x+y)=(ap-bq)(ap+bq)>p+q$ for some integers $a$ and $b$ . So the answer clearly follows

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#8 h May 18, 2025, 6:40 AM

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Notice that each prime sums with $n$ to a distinct perfect square.
So let $k \in N$ such that $k^2>n$
We have, $(k+3)^2-n<\frac{n}{2}$ ,because the max prime is $< \frac{n}{2}$
$k^2+9+6k<\frac{3n}{2}$ $\implies$ $n^2<k<\frac{\frac{n}{2}-9}{6}$ ,now just solve this equation for $n$ you will get $-2<n<2$ ,So you it also does not work.

So it has clearly at most 2 prime factors for $n=p.q>1$ which respectively sums to $n+p=k^2$ and $n+q=(k+1)^2$ .
hence one has to be even = $2$ and the other must be $2k+3$ and $n=k^2-2$

Now just put all in that equation you will get : $k^2-4k-8=0$ no integer solution .
(I don't why my solution was so basic:(

This post has been edited 1 time. Last edited by Mathgloggers, May 18, 2025, 6:41 AM
Reason: m

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#9 h May 18, 2025, 9:28 PM

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mohsen wrote:

Find all positive integers n>2 such that sum of n and any of its prime divisors is a perfect square.

Here is my solution that I actually used in contest
Trivialy there exist positive intiger m and non negative integer q so that n=m^2+q and n<(m+1)^2 so q<2m+1 now there are 2 cases
Case 1: n has at least two different prime divisor
Let p_1 and p_2 be prime divisor of n and WLOG p_2>p_1 notice there exist intiger r_1 and r_2 such that n+p_1=(m+r_1)^2 and n+p_2=(m+r_2)^2 and p_2 divide n therfore p_2 divide n+p_2 so p_2 divide (m+r_2)^2 thus p_2 divide m+r_2 but we know p_2=(m+r_2)^2-n=m^2+2(r_2)m+(r_2)^2-m^2-q and we know any natural divisor of a natural number is smaller or equal to that number so we have
2(r_2)m +(r_2)^2-q =< m+r_2
But we have r_2=>2 and q<=2m+1 thus contradiction is trivial
Case 2: n has only one prime factor
Thus there exist prime number p and positive intiger a such that n=p^a therfore p^a+p is square but for any a larger then 1 we know that p divide n and p^2 do not divide n thus n can't be a square so a=1 and for a=1 we would get 2p is square thus 4 divide 2p and p is even thus p=2

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