Showing $\sum_{m=1}^{q-1}m^{(q+1)/2}\equiv0\bmod{q}$, where $q$ is an odd prime congruent to $3 \bmod{4}$

Question

I'm reading through Davenport's "Multiplicative Number Theory", and came across this expression on page 53. $$ \sum_{m=1}^{q-1} m^{(q+1)/2} \equiv 0 \mod q, $$ where $q$ is an odd prime congruent to $3\bmod 4$.

In the book, this is stated as if it should be immediately clear, but for some reason, I just can't see it. Can anyone help with my tunnel vision?

Answer 1

Assume $q>3$, but there is no need to set a value for $q\pmod 4$. We remark that the claim is false for $q=3$.

Let $t$ be a non-zero residue $\pmod q$ such that $t^{(q+1)/2}\not \equiv 1 \pmod q$. Such a $t$ exists since the order of any $t$ for which $t^{(q+1)/2} \equiv 1 \pmod q$ would have to be a divisor of $\gcd(q-1, \frac {q+1}2)≤2$ and, since $q>3$, we can avoid such $t$.

Now, multiplying the list of residues by $t$ just permutes them. It follows that $$\sum m^{(q+1)/2}\equiv t^{(q+1)/2}\sum m^{(q+1)/2}\implies \sum m^{(q+1)/2}\equiv 0 \pmod q$$ and we are done.

Answer 2

In the case $q \equiv 1 \pmod{4}$ we have that $q-1$ and $\frac{q+1}{2}$ are coprime so raising numbers to the $\frac{q+1}{2}$ power induces a bijection on the unit group.

If $q \equiv 3 \pmod{4}$ this is not necessarily true since you get the sum of the quadratic residues (take $q = 3$).

More generally, you could write this as a geometric series by considering a generator of the unit group.