If $p$ is the prime of the form $4k+1$. Prove that $(1/p)+(2/p)+(3/p)+\ldots+ (P/p) = 0$

Question

$p$ is the prime of the form $4k+1$. Prove that: $$(1/p)+(2/p)+(3/p)+\ldots+ (P/p) = 0$$

$P=(p-1)/2.$

Answer 1

If $p\equiv 1\pmod 4$, then $\left(\frac{-1}p\right)=1$, hence $$ \sum_{k=1}^{p-1}\left(\frac kp\right)=\sum_{k=1}^{(p-1)/2}\left(\left(\frac kp\right)+\left(\frac {-k}p\right)\right)=2\sum_{k=1}^{(p-1)/2}\left(\frac kp\right).$$ But as there are just as many squares as non-squares, the sum on the left is zero.

Answer 2

Is there something wrong, for every item, for example, $3/p$ is greater than 0?

It is $$\frac{(1+2k)k}{4k+1}$$