Law of Quadratic Reciprocity Equivalent Statement 
Let $p,q$ be two distinct odd primes. Then $(\frac{q}p)=1 \iff p=\pm\beta^2 \pmod{4q}$ for some odd $\beta$. Show that this statement is eqivalent to the Law of Quadratic Reciprocity.

I'm trying to grapple with what the question is actually asking me to show. 
Do I split into various cases of what $p$ and $q$ could possibly be (ie. $1 \pmod 4$ and $3 \pmod 4$) and then show that in each case, the statement holds?

We do one of the four cases. Because $p$ and $q$ both of the shape $4k+1$ is "too easy" and does not fully illustrate the problems we can bump into, we deal with the case  $p$ of the form $4k+3$ and $q$ of the form $4k+1$. 
Suppose that $(q/p)=1$, with $p$ of the form $4k+3$ and $q$ of the form $4k+1$. We want to show that $p\equiv \pm \beta^2\pmod{4q}$ for some odd $\beta$.
Note that by Quadratic Reciprocity we have $(p/q)=1$. So $p$ is a quadratic residue modulo $q$. This means that $p\equiv \alpha^2\pmod{q}$ for some $\alpha$.  But $-1$ is a quadratic residue of $q$, since $q$ is of the form $4k+1$. So $-1\equiv \gamma^2\pmod{q}$ for some $\gamma$, and therefore 
$$p\equiv -(\alpha\gamma)^2\pmod{q}.$$ 
Without loss of generality we may assume that $\alpha\gamma$ is odd. If it isn't,  replace it by $q-\alpha\gamma$.
Since the square of an odd number is congruent to $1$ modulo $4$, we have  $$p\equiv -(\alpha\gamma)^2\pmod{4}.$$
It follows that $p\equiv -(\alpha\gamma)^2\pmod{4q}$.   
The reverse direction is straightforward. Reverse directions are not really needed if we deal with the "forward" direction in all four cases. 
