For each odd prime p, we consider the two numbers: A = sum of all 1 <= a < p suc

ID: 2983573 • Letter: F

Question

For each odd prime p, we consider the two numbers:
A = sum of all 1 <= a < p such that a is a quadratic residue modulo p

B = sum of all 1 <= a < p such that a is a nonresidue modulo p

I did some work and saw that the primes p < 100 with A = B are {5, 13, 17, 29, 37, 41, 53, 61, 73, 89, 97}.

I ned to prove that my criterion that: if p is a such a prime such that p is congruent to 1 (mod 4), then A = B.

I don't understand how to use the fact that -1 is a quadratic residue to solve this...

p is a prime congruent to 1 (mod 4)

=> -1 is a quadratic residue mod p

=> (i) and (ii)

where (i) a is a quadratic residue iff p-a is is

(ii) b is a quadratic nonresidue iff p-b is is

This follows because p-c = (-1)c modulo p and -1 is a square modulo p

(p-c/p) = (-c/p) = (-1/p) (c/p) = (c/p)

We also know the fact that number of quadratic residues = number of nonquadratic residues for odd primes

So in the summation A we (p-1)/2 elements

This along with property (i) implies

A = [(p-1)/2*p]/2 (since A = (p-1)/2 * p -A as summation x_i = summation p - x_i )

But summation of all numbers from 1 to p-1

is (p-1)p/2

Hence A = B = p(p-1)/4

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