I am trying to determine which of the prime numbers $2$, $3$, $5$, $7$, $11$, $13$, $17$, $19$, does $x^{2}+1$ have a solution in the $p$-adic numbers $\mathbb{Q}_{p}$.
I know that this is the same as asking "for which primes does $\mathbb{Q}_{p}$ have $i$ such that $i^{2}=-1$?" but I don't know how to answer that.
I should add that I have found that... $i\in\mathbb{Q}_{p}$ with $i^{2}=-1$ if and only if $p\equiv1\bmod{4}$. So obviously that shortens my list a bit. But how would I do the question without this result?
This question has been asked many times here, I’m sure, but here’s a slightly different answer.
First, if $p\not\equiv1\pmod4$, then there is no $i$ in $\Bbb Q_p$, ’cause there’s no square root of $-1$ in the prime field $\Bbb F_p$.
On the other hand if $p\equiv1\pmod4$, the multiplicative group $\Bbb F_p^\times$ is cyclic of order $p-1$, divisible by $4$, so that there is $r\in\Bbb Z$ with $r^2\equiv-1\pmod p$. Now consider the set of all elements of $\Bbb Z_p$, the $p$-adic integers, congruent to this $r$ modulo $p$. It’s compact, and in it, the map $x\mapsto x^p$ is contractive. (You’ll need to prove this, but it’s easy.) By Banach Fixed Point Theorem, the map has a fixed point $r_0$, which is therefore a root of unity with $r_0^2\equiv-1\pmod p$, thus a square root of $-1$ in $\Bbb Z_p\subset\Bbb Q_p$.