Equivalent statement to Quadratic reciprocity

Question

One corollary of quadratic reciprocity is that $(\frac{p}q)$ depends only on the equivalence class of $q$ modulo $4p$. Is this enough to derive quadratic reciprocity. While equivalent does not really make sense in this context since we only have one model for the integers, I am looking for an "easy" way to derive quadratic reciprocity from the statement.

Answer 1

Your statement that "$\left(\frac pq\right)$ depends only on the equivalence class of $q$ modulo $p$" is not correct. Quadratic reciprocity relates $\left(\frac pq\right)$ and $\left(\frac qp\right)$ with a power of $-1$, namely $(-1)^{\frac{p-1}{2}\frac{q-1}{2}}$, that depends on the actual value of $q$ and not just on its residue class modulo $p$. $\left(\frac qp\right)$ depends on just $q$'s residue class but $\frac{q-1}{2}$ needs more than that.

If you are "looking for an 'easy' way to derive quadratic reciprocity" remember that it was developed by multiple mathematicians, including the greatest mathematician of all time, J.C.F. Gauss. Gauss considered it to be his greatest work, so I very much doubt that you will find an "easy" way to prove it. Easier, perhaps, but not easy. The Wikipedia link states that over $200$ proofs have been published.