Sum of Legendre symbol $\left(\frac{n^2-a}{p}\right)$ (More explanation)

Question

In fact, there are several same questions, but I still post it here:

If $(a,p)=1$, $p$ an odd prime, then $\sum_{n=1}^{p}\left(\frac{n^2+a}{p}\right)=-1$.

In those same posts, I tried to read the proof https://artofproblemsolving.com/community/c146h150500p849406 and Number Theory: Solutions of $ax^2+by^2\equiv1 \pmod p$. But I cannot understand the full proof. Instead, I was stuck in some way. Right now, I am trying to use "Eulers' Criterion" to do this question: $\left(\frac{n^2+a}{p}\right)\equiv (n^2+a)^{\frac{p-1}{2}}$ (mod $p$), thus: $\sum_{n=1}^{p}\left(\frac{n^2+a}{p}\right)\equiv \sum_{n=1}^{p}(n^2+a)^{\frac{p-1}{2}}$ (mod $p$).

Then, I tried to apply the binomial expansion to further simplify. And that is the step where I was stuck. Can anyone further explain the mechanism?

Do you know the formula for $1^k + 2^k + \cdots + p^k$ modulo $p$ when $k$ is an integer between $0$ and $p-1$ ? That said, you will need some care in your approach: If you prove that the sum is $\equiv -1 \mod p$, it won't directly follow that it is $= -1$ as an integer (it could be $p-1$, too). — darij grinberg, Nov 15 '18 at 15:18
Also, it is worth specifying where exactly you're stuck, e.g. in comments under the respective answers. — darij grinberg, Nov 15 '18 at 15:20
No. But i am searching and reading after your comment. I saw this in one of the proof. So, this is important in this question? — Jason Ng, Nov 15 '18 at 15:20
Yes, it is. See https://math.stackexchange.com/questions/433678/sums-of-powers-below-a-prime . — darij grinberg, Nov 15 '18 at 15:22
When I expand $(n^2+a)^{p-1/2}$, I know the first term is $n^{p-1}$, which is 1 modulo $p$. But the rest of the terms I don't know how to simplify under modulo $p$. — Jason Ng, Nov 15 '18 at 15:22
You can't simplify the single terms, but you can simplify their sums over all $n$. — darij grinberg, Nov 15 '18 at 15:23

Sum of Legendre symbol $\left(\frac{n^2-a}{p}\right)$ (More explanation)

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