Solution: Here 'p ' is an odd prime number such that and there are (p-1) / 2 quadratic residues of p.
As ( If a1 is quadratic residue of p then p - a1 = ap-1/2 ) so,
We will see it with an example,
Let p = 13 and 1,3,4,9,10 and 12 are quadratic residues of p=13.
We can see that number of quadratic residues = p-1 /2 = 13-1 /2 = 6 residues (1,3,4,9,10 and 12 )
Now 1+12 = 3+10 = 4+9 = 13
i.e 1 + 3 + 4 +9 + 10 + 12 = (1+12)+(3+10)+(4+9) = 13+13+13
as p-1 / 4 = 13-1 / 4 = 3 so 3 pairs of 13.
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