$\sum_{t=1}^{p-1} (t(1-t))^{\frac{p-1}{4}}$

Question

$\sum_{t=1}^{p-1} (t(1-t))^{\frac{p-1}{4}} \equiv 0$ mod $p$

Here $p \equiv 1$ mod $4$ and prime. I don't have any idea about how to prove this statement.

I checked that for $p=5,p=13,p=17$ and also checked that $p=9$ (not prime). In $p=9$ case, the sum is not zero but 3, so I think the sum is zero divisor in $\mathbb{Z}_p$ but I couldn't show my claim.

Answer 1

Althoght I didn't prove that the sum in zero divisor in $\mathbb{Z}_p$ but in the case of p is a prime, $$ \sum_{t=1}^{p-1}(t-t^2)^{\frac{p-1}{4}} $$

is a polynomial whose degree less than $\frac{p+1}{2}$ and each term (up to degree) has same coefficients when $t$ runs over $1$ to $p-1$ so the sum is zero in mod$p$ by the following lemma.

Lemma. $1^k+2^k+\cdots+(p-1)^k \equiv 0$ mod $p$ if $p-1$ does not divide $k$.

(This idea is not mine. It's from my university colleague.)