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.
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.