Square in a finite field

Question

I need to prove: $a$ is a square in $\mathbb{Z}$ iff $a\mod p$ is a square in $\mathbb{F_p}$ for every $p$ prime. First side ($\rightarrow$) is trivial and is derived directly from multiplicativity of modulo. Now I need to prove the other side ($\leftarrow$). I tried taking $p>a$ so $a\equiv a\mod p$, and I know that $a$ is a square so there exists $b<p$ s.t $\space$$b^2\mod p \equiv a$. But, $b$ is not necessarily the root of $a$ in $\mathbb{Z}$, so I'm stuck. I also know that for each $p$, $(\frac{a}{p})=1$ but this doesn't get me anywhere. Any hint would help.

Answer 1

If think the easiest way is to use quadratic reciprocity, that if $a$ is not a square then for $p\ne 2$, $(\frac{a}p) =\chi(p) $ with $\chi$ a non-trivial Dirichlet character modulo $4a$,

there is $m>0$ such that $\chi(m)=-1$, some prime $p|m$ is such that $-1=\chi(p)=(\frac{a}p)$