How to prove this summation involving Legendre symbol: $\sum\limits_{k=1}^{p-1}k\left(\frac{k}{p}\right)=0 $

Question

If $p\equiv 1 \pmod{4}$ is a prime number, prove that $$\sum_{k=1}^{p-1}k\biggl(\dfrac{k}{p}\biggr)=0 $$

Any suggestion how to prove it I will appreciate.

Answer 1

may help to note that $$ (p-k)^2 = p^2 -2kp + k^2 \equiv k^2 \mod{p} $$

Answer 2

For an odd prime $p$, exactly $\frac{p-1}{2}$ of the numbers $1\leq k\leq p-1$ are quadratic residues (apply the first isomorphism theorem to the homomorphism $\mathrm{sq}:(\mathbb{Z}/p\mathbb{Z})^\times\to(\mathbb{Z}/p\mathbb{Z})^\times$ defined by $\mathrm{sq}(x)=x^2$).

For a prime $p\equiv 1\bmod 4$, and any $1\leq k\leq p-1$, $k$ is a quadratic residue iff $p-k$ is a quadratic residue (use the fact that $-1$ is a quadratic residue modulo $p$).

Thus, there are $\frac{p-1}{4}$ pairs $(k,p-k)$ of quadratic residues among the numbers between $1$ and $p-1$, and also $\frac{p-1}{4}$ pairs $(k,p-k)$ of non-quadratic residues.