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Source: 2025 Taiwan TST Round 2 Independent Study 2-N

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#1 h Mar 27, 2025, 5:17 AM • 1 Y

Y by Tintarn

Find all prime number pairs $(p, q)$ such that $p^q+q^p+p+q-5pq$ is a perfect square.

Proposed by chengbilly

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vi144

#2 h Mar 27, 2025, 7:46 AM • 1 Y

Y by Ciobi_

First, suppose $p, q \ge 3$ . Taking $\pmod 4$ we see that $4$ does not divide $p-q$ . WLOG suppose that $p\equiv 1 \pmod 4$ and $q\equiv 3 \pmod 4$ .

Taking $\pmod p$ and $\pmod q$ gives (using FLT) the following relation:
$\left(\frac{2p}{q}\right)=1=\left(\frac{2q}{p}\right)$ from which, using quadratic reciprocity and $p\equiv 1 \pmod 4$ we get
$\left(\frac{2}{p}\right)=\left(\frac{2}{q}\right)$ Thus, modulo $8$ , $(p, q)$ is either $(1, 7)$ or $(5, 3)$ .
Evaluating both these options $\pmod 8$ yields that the expression is $5 \pmod 8$ , which cannot be a perfect square. Thus we must have $p=2$ or $q=2$ . WLOG suppose $p=2$ . We wish to find the primes $q$ such that
$2^q+q^2+2-9q$ is a perfect square.
$q=2, 3$ are not solutions. For $q>3$ , consider the expression $\pmod 3$ :
$2^q+q^2+2-9q\equiv 2^q+1+2\equiv 2 \pmod 3$ as $q$ must be odd. Thus we have no solutions.

This post has been edited 1 time. Last edited by vi144, Mar 27, 2025, 8:20 AM

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#3 h Mar 28, 2025, 3:24 PM • 1 Y

Y by MS_asdfgzxcvb

We show that there are no solutions. Suppose otherwise. First note that $p = q = 2$ fails, so one of $p,q$ is odd. WLOG $p$ is odd. Let the expression be denoted as $n^2$ .

Case 1: $q = 2$ .
Then $p^2 - 9p + 2^p + 2$ is a square. Since $p$ is odd, $2^p \equiv 2 \pmod 3$ . Thus, $n^2 \equiv p^2 + 2 + 2 \equiv p^2 + 4 \pmod 3$ . Since $2$ isn't a QR mod $3$ , $p^2$ cannot be $1 \pmod 3$ , meaning $p = 3$ , which can be checked to fail.

Case 2: $q > 2$ .
Firstly, since $p,q$ are odd, $p^q \equiv p \pmod 8, q^p \equiv q \pmod 8$ .

We note that $n^2\equiv p + q +p + q - 5pq \equiv 2(p + q) - pq \equiv -pq \pmod 4,$ so one of $p,q$ is $1 \pmod 4$ and the other is $3 \pmod 4$ (as $n^2$ has to be $1 \pmod 4$ ), so $8 \mid 2(p + q)$ .

Additionally, by taking modulo $8$ , $n^2 \equiv 2(p + q) - 5pq \equiv -5pq \pmod 8$
Since $1$ is the only odd QR modulo $8$ , we note that $-5pq \equiv 1\pmod 8$ , so $pq \equiv 3 \pmod 8$ .

In particular, this means that $2$ is not a QR modulo $pq$ .

Now, note that $n^2 \equiv 2p \pmod q$ and $2q \pmod p$ , so (using Legendre/Jacobi Symbol) $\left( \frac{2p}{q} \right) = \left( \frac{2q}{p} \right) = 1$
Therefore $1 = \left( \frac{2p}{q} \right) \left( \frac{2q}{p} \right) = \left( \frac 2q \right) \left( \frac pq \right) \left( \frac 2p \right) \left( \frac qp \right) = \left( \frac{2}{pq} \right) \left( \frac pq \right) \left( \frac qp \right) = -1$ absurd (the last part follows from Law of Quadratic Reciprocity).

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