Iran second round 2025-q1

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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 three different primes $p,q,r$ such that $p\equiv q \equiv r \equiv 3 \mod 4$ . Note that $n$ is also square-free.
So we get
$p(\frac{n}{p}+1) = x^2 \Rightarrow (\frac{p}{q})=(\frac{p}{r})$ Similarly $(\frac{q}{p})=(\frac{q}{r})$ and $(\frac{r}{p})=(\frac{r}{q})$ .
Now note that $(\frac{p}{q})(\frac{q}{p})=(-1)^{\frac{p-1}{2}. \frac{q-1}{2}}=-1 \Rightarrow (\frac{p}{q})=-(\frac{q}{p})$ this gives $(\frac{r}{q})=(\frac{q}{r})$ but $(\frac{r}{q})=-(\frac{q}{r})$ which is a contradiction.
A similar approach works for $n$ having two primes of the form $4k-1$ and at least one of the form $4t+1$ (if $n = 2p$ where $p\equiv 3 \mod 4$ , we must have $3p$ is a perfect square which gives $p=3$ which doesn't work. if $n=2pq$ where $p,q \equiv 3 \mod 4$ , we get $p|2q+1,q|2p+1$ which gives ${p,q}={3,7}$ which doesn't work. if $n=p$ we get $p=2$ but $n>2$ and if $n=pq$ we get ${p,q}={2,3}$ which also 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 1 time. Last edited by Parsia--, Apr 22, 2025, 4:06 PM
Reason: Added somethings

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#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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