Prove that $21|a^2 +b^2 \implies 441|a^2+b^2$ 
Prove that $$21|a^2+b^2 \implies 441|a^2+b^2$$

I can show that if $3$ divides $a^2+b^2$ than so does $9$, so the part that's left to do is to prove that $7$ and $49$ divide $a^2+b^2$ and I am stuck at this point.

Note that
\begin{align}
1^2&\equiv 1\pmod{7},\\
2^2&\equiv 4\pmod{7},\\
3^2&\equiv 2\pmod{7},\\
4^2&\equiv 2\pmod{7},\\
5^2&\equiv 4\pmod{7},\\
6^2&\equiv 1\pmod{7}.
\end{align}
It is impossible to get 2 numbers with replacement from $\{1,2,4\}$ to get a sum that is divisible by $7$. Thus, to have $a^2+b^2\equiv 0\pmod{7}$, it must be that $a,b\equiv 0\pmod{7}$ and so $a^2,b^2\equiv 0\pmod{49}$.
