Q:

Proving statement with Gauss's reciprocity law

State Gauss' reciprocity law for two distinct odd primes $p$ and $q$.
  Hence, given an odd prime $p$, prove carefully that the following
  statements are equivalent:

$p\neq5$ and there is an intenger $k$ such that $p$ divides $5-k^2$;
$p$ is congruent to $\pm1\mod{5}$.

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I am asked to solve the above question.  I have got the Gauss's reciprocity law below. 

For distinct odd primes $p$ and $q$, the theorem states:
  $$\operatorname{Leg}(q,p)=\begin{cases}\operatorname{Leg}(p,q)&\text{if
> $p\equiv1$ or $q\equiv1\mod4$},\\-\operatorname{Leg}(p,q)&\text{if
> $p\equiv-1$ and $q\equiv-1\mod4$}.\end{cases}$$

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I am stuck when I try to do the second bit.  Can anyone give me some hints on how to start with the first statement? 

A:

The first statement is telling you that there is an integer $k$ such that $p|5-k^2$, this is equivalent to say that there is a $k$ such that $k^2\equiv 5\mod p$, i.e. is equivalent to say that $5$ is a quadratic residue mod $p$, i.e. equivalent to $Leg(5,p)=1$.
So, what you are trying to prove is $Leg(5,p)=1\iff p\equiv \pm 1\mod 5$. You should prove this using quadratic reciprocity.  Do you see how?