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?

Just to show that I'm on the right track, I'll try solving the easier case and hopefully somebody can check it for me.
Suppose that $(q/p)=1$, with $p$ of the form $4k+1$ 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 $q$. 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 $p\equiv (\alpha\gamma)^2\pmod{4q}$.
(I suppose we can just take $\gamma=1$ here?)  
