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.


Lemma. If a prime $p \equiv 3\pmod{4}$ and $p$ divides $a^2+b^2$ then $p$ divides $a,b$.

Proof. Assume the contrary that $p \nmid a,p \nmid b$.
By Fermat's little theorem, we have $a^{p-1} \equiv 1 \pmod{p}, b^{p-1} \equiv 1 \pmod{p}$ so $a^{p-1}+b^{p-1} \equiv 2 \pmod{p}$.
On the other hand, note that $x+y$ divides $x^{2k+1}+y^{2k+1}$ for all integer $x,y,k$ and $k \ge 0$. Hence,since $p \equiv 3 \pmod{4}$ so $(p-1)/2$ is odd, we obtain $a^2+b^2$ divides $(a^2)^{(p-1)/2}+(b^2)^{(p-1)/2}$. Since $p \mid a^2+b^2$ so $p$ divides $(a^2)^{(p-1)/2}+(b^2)^{(p-1)/2}=a^{p-1}+b^{p-1}$, a contradiction since $a^{p-1}+b^{p-1}\equiv 2 \pmod{p}$ and $p \ge 3$.  $\square$

Back to the problem, since $3,7 \equiv 3 \pmod{4}$ are primes so if $7 \mid a^2+b^2$ then $7 \mid a, 7 \mid b$ which means $7^2 \mid a^2+b^2$. Similarly, $3^2 \mid a^2+b^2$. Thus, $441 \mid a^2+b^2$.
