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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.

Ng Chung Tak
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2 Answers2

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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 $a^2,b^2\equiv 0\pmod{49}$.

yurnero
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  • These are squares (modulo $7$), so one can even save oneself the (trivial) hassle of computing the squares of $4, 5$, and $6$ (modulo $7$), since those are equal to the squares of $-3, -2$, and $-1$ (modulo $7$). – Brian Tung Nov 18 '16 at 19:14
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By considering the ring of gaussian integers $\Bbb Z[i]$, a prime $p\in\Bbb N$ remains a prime in $\Bbb Z[i]$ if and only if $p\equiv -1\pmod 4$. Such is the case for $3$ and $7$. However in $\Bbb Z[i]$ it holds $a^2+b^2=(a+ib)(a-ib)$.

So, if, say, $7\mid a^2+b^2$, then $7$ divides either $a+ib$ or $a-ib$ in $\Bbb Z[i]$. But, by the identity $\overline{7z}=\overline 7\overline z=7\overline z$, we know that $7$ divides either of those numbers if and only if it divides both of them. So $49\mid (a+ib)(a-ib)=a^2+b^2$.