Let p be a prime number with p>3. Prove that the sum of the quadratic residues modulo p is divisible by p.

Originally Posted by babygill
Let p be a prime number with p>3. Prove that the sum of the quadratic residues modulo p is divisible by p.
How far have you gotten with this?

-Dan

I don't know where to start

Okay. I've refreshed my memory of this topic, but I am in no way a professional.

First, I am thinking that your problem statement is incorrect. (Which may mean that I'm incorrect, but anyway...)

For example, take a look at the multiplicative group mod 7: $\mathBB{Z} / _7 \mathbb{Z} = \{ 1, 2, 3, 4, 5, 6 \}$.

Squaring each of these elements gives the quadratic residues by inspection: q = 1, 4. The sum of these is obviously not divisible by 7. I think what you are looking for is the sum of the square of all members of the group. This can be easily shown by looking at the summation:
$\sum_{i = 1}^6~i^2$

This turns out to be divisible by 7 as required. (And thus the sum of the residues is also divisible by 7. Prove this!) I leave it to you to look at the general case.

-Dan