2023 SSMO Team Round Problems/Problem 14

Problem

Find the sum of all perfect squares of the form $2p^3 - 5p^2q + q^2$ where $p$ and $q$ are positive integers such $p$ is prime and $p \nmid q$ .

Solution

If $p = 2$ , the expression becomes $(q - 10)^2 - 84,$ which can equal either $400$ or $16$ .

If $p > 2$ , then let $2p^3 - 5p^2q + q^2 = k^2.$

We rearrange and factor as $p^2(2p - 5q) = (k + q)(k - q).$

Note that $p$ does not divide $\gcd(k + q, k - q) = \gcd(k + q, 2q)$ , so $p$ cannot divide both factors evenly. Thus, either $p^2 \mid (k + q)$ or $p^2 \mid (k - q)$ , implying that $k = ap^2 \pm q$ for some integer $a$ , and so $k \ge p^2 - q.$

However, this leads to a contradiction: $2p^3 - 5p^2q + q^2 < (p^2 - q)^2 = p^4 - 2p^2q + q^2.$

This means no perfect square is possible in this form for $p > 2$ , so the only valid values come from the $p = 2$ case, where $400 + 16 = \boxed{416}.$

~SMO_Team

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2023 SSMO Team Round Problems/Problem 14

Problem

Solution