Let $f(x)=x^2-a \in \mathbb{Z}[x]$. $$p \in \mathbb{P}, p \neq 2, p^2 \nmid a$$
The equation $f(x)=0$
- If $p \mid a $, the equation has no solution in $\mathbb{Q}_p$
Let $p \nmid a$.
- The equation has exactly two solutions in $\mathbb{Q}_p \Leftrightarrow \left( \frac{a}{p}\right)=1$
- It has no solution in $\mathbb{Q}_p \Leftrightarrow \left( \frac{a}{p}\right)=-1$
Why does it have to hold that $p^2 \nmid a$?
In order to prove the first proposition I thought that since $p \mid a $ and $p^2 \nmid a$ it has to hold that $a=kp, p \nmid k$. Since $p$ is a prime, it will hold that $(p,k)=1$. Thus $a$ cannot be written as a square and so if $p \mid a$, the equation $x^2-a$ has no solution in $\mathbb{Q}_p$. Is it right?
The second and the third proposition hold from the definition of the Legendre symbol. Am I right?
EDIT: And how could we show that $f(x)=0$ has a solution in $\mathbb{Q}_2 \iff a \equiv 1 \mod 8$ ?
For the first time, it is an application of Eisenstein's criterion.
For the second part, it is more or less correct, but you should consider that the Legendre's symbol is defined for integers, and $\mathbb{Z}_p$ is a subset of $\mathbb{Q}_p$: if $(x/y)^2 \equiv a \pmod{p}$ then $p$ cannot divide $y$, so there exists a multiplicative inverse $z\neq 0$ such that $p\mid yz-1$. It follows that $(xz)^2 \equiv a \pmod{p}$, and now we are talking about integers.