National Math Olympiad

First, let $a$, $b$ and $c$ be even: $a=2a_1$, $b=2b_1$, $c=2c_1$. Then the number $$a^2+b^2+c^2=4(a_1^2+b_1^2+c_1^2)$$ is divisible by $4$.

Now, let us prove the converse. Assume that $a^2+b^2+c^2$ is divisible by $4$. If exactly one of the numbers $a$, $b$ and $c$ were odd or if all three were odd, then the sum $a^2+b^2+c^2$ would be odd. This is not the case.

Finally, assume that exactly two of the numbers $a$, $b$ and $c$ are odd. We may assume that $a$ and $b$ are odd and $c$ is even. Let $a=2a_1-1$, $b=2b_1-1$ and $c=2c_1$ for some positive integers $a_1$, $b_1$ and $c_1$. In this case the number $$a^2+b^2+c^2=(2a_1-1{)}^2+(2b_1-1{)}^2+(2c_1{)}^2=4(a_1^2-a_1+b_1^2-b_1+c_1^2)-2$$ is not divisible by $4$. This contradicts the assumption. The only remaining possibility is that all three numbers are even.