Romanian Mathematical Olympiad

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.