75th Romanian Mathematical Olympiad

From the identity $a^2{b}^2+b^2{c}^2+c^2{a}^2=(ab+bc+ca{)}^2-2abc(a+b+c)$ 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.