All squares modulo p where p is 3 modulo 4 are 4 powers I would like to prove the lemma:

For every prime $p$ such that $p=3 \:(mod\:4)$ if $a$ is square modulo $p$ (a.k.a there exists a number $b$ such that $a=b^2\:(mod\:p)$), then its 4 power modulo $p$.

But I don't have any lead.
Thanks!
 A: The key thing here, one way or another, is to note that $-1$ is not a square modulo $p$. This means that if $b^2=a$ either $b$ or $-b$ is a square. I won't fill in the details because you said you wanted a lead.
A: The following proof is excerpted from one of my answers to an older question . . .

Assume $p$ is a prime, with $p \equiv -1\;\,(\text{mod}\;4)$.

Then $-1$ is not a quadratic residue, mod $p$.

Claim every square in $Z_p$ is also a $4$-th power.

Let $a,b \in Z_p^{*}$, and suppose $a^4=b^4$ in $Z_p$.
\begin{align*}
\text{Then in $Z_p$,}\;\;&a^4=b^4\\[4pt]
\iff\;&a^2=\pm b^2&&\text{[since $Z_p$ is a field]}\\[4pt]
\iff\;&a^2=b^2&&\text{[since $-1$ is not a quadratic residue, mod $p$]}\\[4pt]
\iff\;&a= \pm b&&\text{[since $Z_p$ is a field]}\\[4pt]
\end{align*}
Noting that $b\ne -b$ in $Z_p^*$ (since $p$ is odd), it follows that the map $Z_p^{*} \to Z_p^{*}$ given by $x \mapsto x^4$ is exactly two-to-one, hence the set of $4$-th powers in $Z_p^{*}$ has cardinality $\frac {p-1}{2}$.

But the set of squares in $Z_p^{*}$ also has cardinality $\frac {p-1}{2}$ (since $p$ is prime).

Of course, every $4$-th power is also a square, hence, since the cardinalities are the same, the set of $4$-th powers in $Z_p^{*}$ is the same as the set of squares in $Z_p^{*}$.

Noting that $0$ is both a square and a $4$-th power, it follows, as claimed, that every square in $Z_p$ is also a $4$-th power.
