amparvardi wrote: Let $a,b,c$ be positive integers. Prove that $a^2+b^2+c^2$ is divisible by $4$, if and only if $a,b,c$ are even. If $a,b,c$ are even then indeed $a^2 +b^2 +c^2 \equiv 0 \pmod 4$. Now assume that $a^2 +b^2 +c^2 $ is divisible by $4$, but not all the numbers are even. Then we'd have two of the numbers being odd, say $a$ and $b$. Let $a=2k+1$, $b=2q+1$ and $c=2p$, then we'd have: $a^2 +b^2 +c^2=4k^2 +4k +1 +4q^2 +4q +1 +4p^2 $ which is clearly not divisible by $4$. Hence contradiction.

Here is the solution for those who usually don't study modular arithmetic in High School curriculum. Let $a=2p_1+r_1,\ b=2p_2+r_2,\ c=2p_3+r_3$ with $r_i=0,\ 1\ (i=1,\ 2,\ 3)$, we have $a^2+b^2+c^2=4(p_1^2+p_2^2+p_3^2+p_1r_1+p_2r_2+p_3r_3)+r_1^2+r_2^2+r_3^2.$ $\therefore 4|a^2+b^2+c^2\Longleftrightarrow r_1^2+r_2^2+r_3^2=0\Longleftrightarrow r_1=r_2=r_3=0.$

if not all of them were even then two of them must be odd . Without loss of generality assume that $a,b$ are odd then $a\equiv 1 \pmod 4$ or $a\equiv3 \pmod 4$ therefore $a^{2}\equiv 1 \pmod 4$. (similar we get $b^{2}\equiv 1 \pmod 4$) Hence we get $ a^2+b^2+c^2\equiv 2\pmod 4 $ a contradiction Ofcourse we would get $ a^2+b^2+c^2\equiv 0\pmod 4 $ if all numbers were even