How to prove $x^2=-1$ has a solution in $\mathbb{Q}_p$ iff $p=1\mod 4$

Question

Let $p$ be prime and let $\mathbb{Q}_p$ denote the field of $p$-adic numbers. Is there an elementary way to prove $x^2=-1$ has a solution in $\mathbb{Q}_p$ iff $p=1\mod 4$?

I need this result, but I cannot find a reference. Can some recommend a good book or a set of (easily available) lecture notes to help me out?

Answer 1

Following Euler Criterion

http://en.wikipedia.org/wiki/Euler%27s_criterion

$-1$ is a quadratic residue if $(-1)^{\frac{(p-1)}2}=1$

which happens only when $\frac{(p-1)}2$ is even

so $p-1$ is divisible by 4

so $p\equiv1 (mod4)$

the book i would follow for number theory is Introduction to number theory by Hua Loo Keng

this is a fairly comprehensive book can a bit a too much for elementary course so depending on your syllabus you may want to use an elementary text like Elementary number theory by Rosen.

Answer 2

Are you familiar about number theory? This is equivalent to saying that $x^2 \equiv -1$ in $\mathbb{F}_p = \mathbb{Z}/p\mathbb{Z}$ iff $p \equiv 1$ (mod $4$). But this is a basic fact on number theory, since \begin{equation*} \Big( \frac{-1}{p} \Big) = (-1)^{\frac{p-1}{2}} = 1 \text{ iff } p \equiv 1 (\text{ mod }4) \end{equation*}