2022 USAJMO Problems/Problem 5

Problem

Find all pairs of primes $(p,q)$ for which $p-q$ and $pq-q$ are both perfect squares.

Solution

Let $p-q = a^2$ , $pq - q = b^2$ , where $a, b$ are positive integers. $b^2 - a^2 = pq - q - (p-q) = pq -p$ . So, $b^2 - a^2 = p(q-1) \tag{1}$

$\bullet$ For $q=2$ , $p = b^2 - a^2 = (b-a)(b+a)$ . Then $b-a=1$ and $b+a=p$ . $a=\dfrac{p-1}{2}$ and $p-q = a^2$ . Thus, $p - 2 = \left( \dfrac{p-1}{2} \right)^2 \implies p^2 - 6p + 9 = 0$ and we find $p=3$ . Hence $(p,q) = (3,2)$ .

$\bullet$ For $q=4k+3$ , ( $k\geq 0$ integer), by $(1)$ , $p(4k+2) = b^2 - a^2$ . Let's examine in $\mod 4$ , $b^2 - a^2 \equiv 2 \pmod{4}$ . But we know that $b^2 - a^2 \equiv 0, 1 \text{ or } 3 \pmod{4}$ . This is a contradiction and no solution for $q = 4k + 3$ .

$\bullet$ For $q=4k+1$ , ( $k > 0$ integer), by $(1)$ , $p(4k) = b^2 - a^2$ . Let $k=m\cdot n$ , where $m\geq n \geq 1$ and $m, n$ are integers. Since $p>q$ , we see $p>4k$ . Thus, by $(1)$ , $(b-a)(b+a) = 4p\cdot m \cdot n$ . $b-a$ and $b+a$ are same parity and $4p\cdot m \cdot n$ is even integer. So, $b-a$ and $b+a$ are both even integers. Therefore,

$\left\{ \begin{array}{rcr} b+a = & 2pn \\ b-a = & 2m \end{array} \right.$ or $\left\{ \begin{array}{rcr} b+a = & 2pm \\ b-a = & 2n \end{array} \right.$ Therefore, $a=pn - m$ or $a = pm - n$ . For each case, $p-q = p - 4mn - 1 < a$ . But $p-q = a^2$ , this gives a contradiction. No solution for $q = 4k + 1$ .

We conclude that the only solution is $(p,q) = (3,2)$ .

(Lokman GÖKÇE)

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2022 USAJMO Problems/Problem 5

Problem

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