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?

This an approach through which you have not to split into cases, but have to be more careful.
The canonical, or traditional, reciprocity law is: Let $p, q$ be primes. Then $$(\frac{q}p)=(\frac{p}q)(\frac{-1}q)^{\frac{p-1}{2}}$$.
Now, if $p\equiv \pm \beta^2\pmod {4q}$, then $(\frac{q}p)=(\frac{\pm1}q)(\frac{-1}q)^{\frac{p-1}{2}}$, where the sign is determined by $\pm p\equiv 1\pmod 4$, i.e. $\pm1=(-1)^{\frac{p-1}{2}}$. Hence it follows that $(\frac{q}p)=1$. Conversely, if $(\frac{q}p)=1$, then $(\frac{p}q)=(\frac{-1}q)^{\frac{p-1}{2}}$, so $(\frac{(-1)^{\frac{p-1}{2}}p}q)=1$, namely, $p\equiv\pm\beta^2\pmod {4q}$.
Also, we have to travel back in this directin. That is, we have to show that, if our statement holds, then so does the traditional statement. So, let us see what our statement means: $p\equiv \pm\beta^2\pmod{4q}$ means that $(-1)^{\frac{p-1}{2}}p\equiv \beta^2 \pmod{q}$, or, equivalently, $(\frac{(-1)^{\frac{p-1}{2}}p}q)=1$. But his is nothing else than our traditional law.  

Conclusion:
  This particular formulation needs barely any errort to verify its equivalence with the classical one, once one realizes how the plus or minus sign in the statement is determined by $p$.    

Barring errors. Thanks in advance.  
