Let $p$ be a prime with $p \equiv 1 \pmod{4}$. Let $a$ be the unique integer such that \[p=a^{2}+b^{2}, \; a \equiv-1 \pmod{4}, \; b \equiv 0 \; \pmod{2}\] Prove that \[\sum^{p-1}_{i=0}\left( \frac{i^{3}+6i^{2}+i }{p}\right) = 2 \left( \frac{2}{p}\right),\] where $\left(\frac{k}{p}\right)$ denotes the Legendre Symbol.