# Also an integer a is a quadratic resisdue modulo n if

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Also, an integer a is a quadratic resisdue modulo n if and only if it is a quadratic residue modulo p i for 1 i r . That completes our investigation of the case where n is an odd positive integer. We shall not investigate the case where n is even, as it is a bit cumbersome, and is not of particular importance. 9.2 The Legendre Symbol For an odd prime p and an integer a with gcd( a, p ) = 1, the Legendre symbol ( a | p ) is defined to be 1 if a is a quadratic residue modulo p , and - 1 otherwise. For completeness, one defines ( a | p ) = 0 if p | a . Theorem 9.4 Let p be an odd prime, and let a, b Z , both not divisible by p . Then 1. ( a | p ) a ( p - 1) / 2 (mod p ) ; in particular, ( - 1 | p ) = ( - 1) ( p - 1) / 2 ; 2. ( a | p )( b | p ) = ( ab | p ) ; 3. a b (mod p ) implies ( a | p ) = ( b | p ) ; 4. (2 | p ) = ( - 1) ( p 2 - 1) / 8 ; 5. if q is an odd prime different from p , then ( p | q )( q | p ) = ( - 1) p - 1 2 q - 1 2 . Part (5) of this theorem is called the Law of Quadratic Reciprocity. Part (1) follows from Theorem 9.1. Part (2) is an immediate cosequence of part (1), and part (3) is clear from the definition. The rest of this section is devoted to a proof of parts (4) and (5) of this theorem. The proof is completely elementary, although a bit technical. The proof we present here is taken almost verbatim from Niven and Zuckerman’s book, An Introduction to the Theory of Numbers . 57

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Theorem 9.5 (Gauss’ Lemma) Let p be an odd prime and a relatively prime to p . Define α j := ja rem p for 1 j ( p - 1) / 2 , and let n be the number of indices j for which α j > p/ 2 . Then ( a | p ) = ( - 1) n . Proof. Let r 1 , . . . , r n denote the α j ’s exceeding p/ 2, and let s 1 , . . . , s k denote the remaining α j ’s. The r i and s i are all distinct and non-zero. We have 0 < p - r i < p/ 2 for 1 i n , and no p - r i is an s j ; indeed, if p - r i = s j , then s j ≡ - r j (mod p ), and writing s j = k 1 a and r j = k 2 a for 1 k 1 , k 2 ( p - 1) / 2, we have k 1 a ≡ - k 2 a (mod p ), which implies k 1 ≡ - k 2 (mod p ), which is impossible. It follows that the sequence of numbers s 1 , . . . , s k , p - r 1 , . . . , p - r n is just a re-ordering of 1 , . . . , ( p - 1) / 2. Then we have (( p - 1) / 2)! s 1 · · · s k ( - r 1 ) · · · ( - r n ) ( - 1) n s 1 · · · s k r 1 · · · r n ( - 1) n (( p - 1) / 2)! a ( p - 1) / 2 (mod p ) , and cancelling the factor (( p - 1) / 2)!, we obtain a ( p - 1) / 2 ( - 1) n (mod p ), and the result follows from the fact that ( a | p ) a ( p - 1) / 2 (mod p ). 2 Theorem 9.6 If p is an odd prime and gcd( a, 2 p ) = 1 , then ( a | p ) = ( - 1) t where t = ( p - 1) / 2 j =1 b ja/p c . Also, (2 | p ) = ( - 1) ( p 2 - 1) / 8 . Proof. Let a be an integer relatively prime to p (not necessarily odd), and let us adopt the same notation as in the proof of Theorem 9.5. Note that ja = p b ja/p c + α j , for 1 j k , so we have ( p - 1) / 2 X j =1 ja = ( p - 1) / 2 X j =1 p b ja/p c + n X j =1 r j + k X j =1 s j . Also, we saw in the proof of Theorem 9.5 that the integers s 1 , . . . , s k , p - r 1 , . . . , p - p n are a re-ordering of 1 , . . . , ( p - 1) / 2, and hence ( p - 1) / 2 X j =1 j = n X j =1 ( p - r j ) + k X j =1 s j = np - n X j =1 r j + k X j =1 s j . Subtracting, we get ( a - 1) ( p - 1) / 2 X j =1 j = p ( p - 1) / 2 X j =1 b ja/p c - n + 2 n X j =1 r j .
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