Let p be a prime number with p>5. Prove that the sum of the squares of the quadratic nonresidues modulo p is divisible by p.

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- November 27th 2012, 08:54 AMbabygillQuadratic Residues
Let p be a prime number with p>5. Prove that the sum of the squares of the quadratic nonresidues modulo p is divisible by p.

- July 11th 2013, 03:59 AMbrosnan123Re: Quadratic Residues
Answer :- f we consider the sum:

∑ i^2, where 1 ≤ i ≤ (p-1)/2 (note: only half of the squared numbers)

It is possible to show (by induction) that this sum can be calculated as

∑ i^2 = (p-1)*p*(p+1)/24

Hence the sum of the quadratic residues can be divided by p. (Why does it suffice to sum the first half of the squared numbers between 1 and p-1?).

On the other hand we have the sum:

∑ k, where 1 ≤ k ≤ p-1 (all residues mod p both quadratic and non-)

This is known as

∑ k = (p-1)*p/2

And is clearly divisible by p. Since the sum of the quadratic residues is part of this sum and is divisible by p then the rest of this sum, namely the quadratic non residues, has to be divisible by p.

and also know about How to Factor a Number. - July 11th 2013, 10:40 AMjohngRe: Quadratic Residues
Hi,

brosnan123 has given a complete proof, almost. At the end he sums all the elements instead of the squares of all the elements. Here is his proof fleshed out a bit:

Attachment 28782