Competition problem — Number theory

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Number theory Difficulty 5.3 AIME, harder Prove it Romania

Determine all the natural numbers $a, b, c$ such that $ab + bc + ca$ is a prime number $p$ and $p$ divides the number $a^2b^2 + b^2c^2 + c^2a^2$ .

Solution

From the identity $a^2b^2 + b^2c^2 + c^2a^2 = (ab+bc+ca)^2 - 2abc(a+b+c)$ it follows that $p$ divides $abc(a+b+c)$ . Because $p$ is a prime number, we get $p \mid a$ , $p \mid b$ , $p \mid c$ or $p \mid (a+b+c)$ .

Since $a, b, c < p$ , the first three cases are impossible. The fourth situation can be true if and only if $a = b = c = 1$ . This case works.

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Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.