A quadratic residue equality proof

Question

Let $p$ be an odd prime, $a\in\mathbb{Z}$. Try to prove $$\sum_{y=0}^{p-1}\Big(\frac{y^2+a}{p}\Big)\equiv-1\pmod{p}\tag{1}$$ I find that when $p|a$ it obviously holds, since $$\sum_{y=0}^{p-1}\Big(\frac{y^2}{p}\Big)=p-1$$ Then for $p\!\!\!\not|a$ case, I checked many examples and I guess the LHS may exacatlly equal to $-1$. Then I make it equivalent to consider giving a proof for the following $$\Big(\frac{a}{p}\Big)+2\sum_{y=1}^{\frac{p-1}{2}}\Big(\frac{y^2+a}{p}\Big)=-1\tag{2}$$ which means that we add a same number to modular $p$'s all quadratic residue. I'm really not sure about my guess $(2)$, does the quadratic residue have such translation-like property? And I do not know how to prove $(1)$, can you give me some ideas?