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representants of quadratic residues∗
pahio†
2013-03-22 3:01:55
Theorem. Let p be a positive odd prime number. Then the integers
2
2
1 , 2 , ...,
p−1
2
2
(1)
constitute a complete representant system of incongruent quadratic residues
modulo p. Accordingly, there are p−1
2 quadratic residues and equally many nonresidues modulo p.
Proof. Firstly, the numbers (1), being squares, are quadratic residues modulo p. Secondly, they are incongruent, because a congruence a2 ≡ b2 (mod p)
would imply
p | a+b or p | a−b,
which is impossible when a and b are different integers among 1, 2, . . . , p−1
2 .
Third, if c is any quadratic residue modulo p, and therefore the congruence
x2 ≡ c (mod p) has a solution x, then x is congruent with one of the numbers
±1, ±2, . . . , ±
p−1
2
which form a reduced residue system modulo p (see absolutely least remainders).
Then x2 and c are congruent with one of the numbers (1).
∗ hRepresentantsOfQuadraticResiduesi
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