Prove $v_p(a^2+b^2)$ is even for $a, b\in\mathbb{Z}_p$ and $p\equiv 3\pmod 4$.

Question

Trying to prove that the $p$-adic valuation of $a^2 + b^2$ is even for $p$-adic integers $a$ and $b$ when $p$ is a $3\bmod 4$ prime. The result for integer $a$ and $b$ is a lemma in elementary number theory, since an integer can be expressed as the sum of integer squares if and only if all $3\bmod 4$ primes which divide it do so to an even power. I suspect the proof that this extends to $a, b\in\mathbb{Z}_p$ has to do with the density of $\mathbb{Z}$ in $\mathbb{Z}_p$ and a property of the $p$-adic valuation, but I'm not quite sure where to go with it. Any advice?

Answer 1

Suppose $v_p(a^2+b^2)=2k+1$ is odd, so $a^2+b^2\equiv 0\pmod{p^{2k+1}}$ but $a^2+b^2\not\equiv 0\pmod{p^{2k+2}}$. Now pick integers $x,y$ with $x\equiv a\pmod{p^{2k+2}}$ and $y\equiv b\pmod{p^{2k+2}}$. Now, we have $x^2+y^2\equiv 0\pmod{p^{2k+1}}$ and $x^2+y^2\not\equiv 0\pmod{p^{2k+2}}$, so $v_p(x^2+y^2)=2k+1$. This is a contradiction.

Answer 2

One way to see this is since $p = 3 \bmod 4$ it implies $i \not\in \mathbb{Q}_p$. Because the extension of the absolute value to $\mathbb{Q}_p(i)$ is unique, it implies $|a+bi|_p=|a-bi|_p$ and so $$|a^2+b^2|_p=|a+bi|_p|a-bi|_p=|a+bi|_p^2$$ which has even valuation because $\mathbb{Q}_p(i)$ is an unramified extension.

Answer 3

You need to avoid $a = b = 0$.

If $a = 0$ or $b = 0$, and the other number is not $0$, then the result is clear. So now assume $a$ and $b$ are both nonzero.

Case 1: $v_p(a) \not= v_p(b)$. Then $v_p(a^2) \not= v_p(b^2)$, so $$ v_p(a^2 + b^2) = \min(v_p(a^2),v_p(b^2)) = 2\min(v_p(a),v_p(b)), $$ which is an even number. We did not use $p \equiv 3 \bmod 4$ here.

Case 2: $v_p(a) = v_p(b)$. Let this common value be $k$,so $a = p^ku$ and $b = p^kv$ where $u$ and $v$ are in $\mathbf Z_p^\times$. Then $$ v_p(a^2 + b^2) = v_p(p^{2k}(u^2 + v^2)) = 2k + v_p(u^2 + v^2). $$ Let's show $v_p(u^2 + v^2) = 0$. If $v_p(u^2 + v^2) \not= 0$ then $v_p(u^2 + v^2) \geq 1$, so $u^2 + v^2 \equiv 0 \bmod p$, which implies $u^2 \equiv -v^2 \bmod p$. Since $v$ is invertible mod $p$, due to being in $\mathbf Z_p^\times$, $-1 \equiv (uv^{-1})^2 \bmod p$, so $-1 \bmod p$ is a square. That contradicts $p \equiv 3 \bmod 4$, so $v_p(u^2 + v^2) = 0$.