primes of the form $4k+3$ and sums of squares

Question

It is well-known that if $p$ is a prime of the form $4k+3$ and $p|x^2+y^2$ then $p|x$ and $p|y$. I forget what is the name of this result, and where can I find a proof (please provide a link).

Answer 1

EDIT: note that your binary quadratic form $x^2 + y^2$ has discriminant $\Delta = -4.$

Somewhere you need to find a proof of the fact that, for prime $q \equiv 3 \pmod 4,$ we always have $$ (-1|q) = -1. $$ For instance, Niven and Zuckerman (and Montgomery) page 132, Theorem 3.1 part (5), says $$ (-1|p) = (-1)^{(p-1)/2}. $$ Same thing in Ireland and Rosen, Proposition 5.1.2, Corollary 3, top of page 52.

Given a binary quadratic form $$ \color{blue}{f(x,y) = a x^2 + b xy+ c y^2} $$ with $a,b,c$ integers. Given its discriminant $$ \color{red}{ \Delta = b^2 - 4 a c}, $$ where we require that $\Delta,$ if non-negative, is not a square (so also $\Delta \neq 0,1$).

Proposition: given an odd prime $q$ such that $q$ does not divide $\Delta$ and, in fact, $$ (\Delta| q) = -1, $$ whenever $$ f(x,y) \equiv 0 \pmod q, $$ then BOTH $$ x \equiv 0 \pmod q, \; \; \; \; \; y \equiv 0 \pmod q. $$

Proof: the integers taken $\pmod q$ form a field; every nonzero element has a multiplicative inverse. If either $a,c$ were divisible by $q,$ we would have $\Delta $ equivalent to $b^2$ mod $q,$ which would cause $(\Delta|q)$ to be $1$ rather than $-1.$ As a result, neither $a$ nor $c$ is divisible by $q.$

Next, $q$ is odd, so that $2$ and $4$ have multiplicative inverses mod $q,$ they are non zero in the field. Put togethaer, $$4a \neq 0 \pmod q$$

Now, complete the square: $$ a x^2 + b xy+ c y^2 \equiv 0 \pmod q $$ if and only if $$ 4a(a x^2 + b xy+ c y^2) \equiv 0 \pmod q, $$ $$ 4a^2 x^2 + 4abxy + 4ac y^2 \equiv 0 \pmod q, $$ $$ 4a^2 x^2 + 4abxy + (b^2 y^2 - b^2 y^2) + 4ac y^2 \equiv 0 \pmod q, $$ $$ 4a^2 x^2 + 4abxy + b^2 y^2 - (b^2 y^2 - 4ac y^2) \equiv 0 \pmod q, $$ $$ (4a^2 x^2 + 4abxy + b^2 y^2) - (b^2 - 4ac) y^2 \equiv 0 \pmod q, $$ $$ (2ax + by)^2 - (b^2 - 4ac) y^2 \equiv 0 \pmod q, $$ $$ (2ax + by)^2 - \Delta y^2 \equiv 0 \pmod q, $$ $$ (2ax + by)^2 \equiv \Delta y^2 \pmod q. $$

Now, ASSUME that $y \neq 0 \pmod q.$ Then $y$ has a multiplicative inverse which we are allowed to call $1/y,$ and we have $$ \frac{(2ax + by)^2}{y^2} \equiv \Delta \pmod q, $$ $$ \left( \frac{2ax + by}{y} \right)^2 \equiv \Delta \pmod q. $$ However, the HYPOTHESIS that $ (\Delta| q) = -1 $ says that this is impossible, thus contradicting the assumption that $y \neq 0 \pmod q.$

So, in fact, $y \equiv 0 \pmod q.$ The original equation now reads $$ a x^2 \equiv 0 \pmod q $$ with the knowledge that $a \neq 0 \pmod q,$ so we also get $$ x \equiv 0 \pmod q, \; \; \; \; \; y \equiv 0 \pmod q. $$ $$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$