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Junior Balkan Mathematical Olympiad

Number theory Difficulty 4.7 AIME Prove it JBMO

Problem:
Let $a$ , $b$ , $p$ , $q$ be positive integers such that $a$ and $b$ are relatively prime, $ab$ is even and $p, q \geq 3$ . Prove that
$2 a^{p} b - 2 a b^{q}$
cannot be a square of an integer number.

Solution

Solution:
Without loss of generality, assume that $a$ is even and consequently $b$ is odd. Let $a = 2 a'$ . Then
$2 a^{p} b - 2 a b^{q} = 4 a' b \left(a^{p-1} - b^{q-1}\right)$
If this is a square, then $a'$ , $b$ and $a^{p-1} - b^{q-1}$ are pairwise coprime.

On the other hand, $a^{p-1}$ is divisible by $4$ and $b^{q-1}$ gives the remainder $1$ when divided by $4$ . It follows that $a^{p-1} - b^{q-1}$ has the form $4k + 3$ , a contradiction.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.