How to prove that $21\mid a^2+b^2\Rightarrow 441 \mid a^2 + b^2$ How to prove that $441 \mid a^2 + b^2$ if it is known that $21 \mid a^2 + b^2$.
I've tried to present $441$ as $21 \cdot 21$, but it is not sufficient.

If $3\mid(a^2+b^2)$ then $3$ divides both $a$ and $b$, since $-1$ is not a quadratic residue $\!\!\pmod{3}$.
The same applies $\!\!\pmod{7}$. If $21\mid(a^2+b^2)$, from the CRT we get that $3$ and $7$ divide both $a$ and $b$, hence $3^2$ and $7^2$ divide both $a^2$ and $b^2$ and $3^2\cdot 7^2\mid (a^2+b^2)$ as wanted.
