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?
 A: 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. 
A: 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.  
A: 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?)  
