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Difference between revisions of "2023 SSMO Team Round Problems/Problem 14"

(Created page with "==Problem== Find the sum of all perfect squares of the form <math>2p^3 - 5p^2q + q^2</math> where <math>p</math> and <math>q</math> are positive integers such <math>p</math> i...")
 
 
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==Solution==
 
==Solution==
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If <math>p = 2</math>, the expression becomes <cmath>(q - 10)^2 - 84,</cmath> which can equal either <math>400</math> or <math>16</math>.
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If <math>p > 2</math>, then let <math>2p^3 - 5p^2q + q^2 = k^2.</math>
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We rearrange and factor as <cmath>p^2(2p - 5q) = (k + q)(k - q).</cmath>
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Note that <math>p</math> does not divide <math>\gcd(k + q, k - q) = \gcd(k + q, 2q)</math>, so <math>p</math> cannot divide both factors evenly. Thus, either <math>p^2 \mid (k + q)</math> or <math>p^2 \mid (k - q)</math>, implying that <cmath>k = ap^2 \pm q</cmath> for some integer <math>a</math>, and so <cmath>k \ge p^2 - q.</cmath>
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However, this leads to a contradiction: <cmath>2p^3 - 5p^2q + q^2 < (p^2 - q)^2 = p^4 - 2p^2q + q^2.</cmath>
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This means no perfect square is possible in this form for <math>p > 2</math>, so the only valid values come from the <math>p = 2</math> case, where <cmath>400 + 16 = \boxed{416}.</cmath>
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~SMO_Team

Latest revision as of 21:38, 9 September 2025

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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