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# For any d p 1 consider the d power map on z p that

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For any d | ( p - 1), consider the d -power map on Z * p that sends α Z * p to α d . The image of this map is the unique subgroup of Z * p of order ( p - 1) /d , and the kernel of this map is the unique subgroup of order d (c.f., Theorem 4.24). This means that the image of the 2-power map is of order ( p - 1) / 2 and must be the same as the kernel of the ( p - 1) / 2-power map. Since the image of the ( p - 1) / 2-power map is of order 2, it must be equal to the subgroup { [ ± 1 mod p ] } . The kernel of the 2-power map is of order 2, and so must also be equal to the subgroup { [ ± 1 mod p ] } . Translating from group-theoretic language to the language of congruences, we have shown: Theorem 9.1 For an odd prime p , the number of quadratic residues a modulo p , with 0 < a < p , is ( p - 1) / 2 . Moreover, if x is a square root of a modulo p , then so is - x , and any square root y of a modulo p satisfies y ≡ ± x (mod p ) . Also, for any integer a 6≡ 0 (mod p ) , we have a ( p - 1) / 2 ± 1 (mod p ) , and moreover, a is a quadratic residue modulo p if and only if a ( p - 1) / 2 1 (mod p ) . Now consider the case where n = p e , where p is an odd prime and e > 1. We also know that Z * p e is a cyclic group of order p e - 1 ( p - 1), and so everything that we said in discussing the case Z * p applies here as well. Thus, for a 6≡ 0 (mod p ), a is a quadratic residue modulo p e if and only if a p e - 1 ( p - 1) / 2 1 (mod p e ). However, we can simplify this a bit. Note that a p e - 1 ( p - 1) / 2 1 (mod p e ) implies a p e - 1 ( p - 1) / 2 1 (mod p ), and by Theorem 4.23 (Fermat’s Little Theorem), this implies a ( p - 1) / 2 1 (mod p ). Conversely, by Theorem 7.7, a ( p - 1) / 2 1 (mod p ) implies a p e - 1 ( p - 1) / 2 1 (mod p e ). Thus, we have shown: Theorem 9.2 For an odd prime p and positive integer e , the number of quadratic residues a modulo p e , with 0 < a < p e , is p e - 1 ( p - 1) / 2 . Moreover, if x is a square root of a modulo p e , then 56

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so is - x , and any square root y of a modulo p e satisfies y ≡ ± x (mod p e ) . Also, for any integer a 6≡ 0 (mod p ) , we have a p e - 1 ( p - 1) / 2 ≡ ± 1 (mod p ) , and moreover, a is a quadratic residue modulo p e iff a p e - 1 ( p - 1) / 2 1 (mod p e ) iff a ( p - 1) / 2 1 (mod p ) iff a is a quadratic residue modulo p . Now consider an arbitary odd positive integer n . Let n = Q r i =1 p e i i be its prime factorization. Recall the group isomorphism implied by the Chinese Remainder Theorem: Z * n = Z * p e 1 1 × · · · × Z * p er r . Now, ( α 1 , . . . , α r ) Z * p e 1 1 × · · · × Z * p er r is a square if and only if there exist β 1 , . . . , β r with β i Z * p e i i and α i = β 2 i for 1 i k , in wich case, we see that the square roots of ( α 1 , . . . , α r ) comprise the 2 r elements ( ± β 1 , . . . , ± β r ). Thus we have: Theorem 9.3 Let n be odd positive integer n with prime factorization n = Q r i =1 p e i i . The number of quadratic residues a modulo n , with 0 < a < n , is φ ( n ) / 2 r . Moreover, if a is a quadratic residue modulo n , then there are precisely 2 r distinct integers x , with 0 < x < n , such that x 2 a (mod n ) .
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