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Unformatted text preview: 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 reordering 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 . Note that ( p 1) / 2 X j =1 j = p 2 1 8 , which implies ( a 1) p 2 1 8 ≡ ( p 1) / 2 X j =1 b ja/p c  n (mod 2) . If a is odd,this implies n ≡ ( p 1) / 2 X j =1 b ja/p c (mod 2) . 58 If a = 2, this — along with the fact that b 2 j/p c = 0 for 1 ≤ j ≤ ( p 1) / 2 — implies n ≡ p 2 1 8 (mod 2) . The theorem now follows from Theorem 9.5. 2 Note that this last theorem proves part (4) of Theorem 9.4. The next theorem proves part (5). Theorem 9.7 If p and q are distinct odd primes, then ( p  q )( q  p ) = ( 1) p 1 2 q 1 2 . Proof. Let S be the set of pairs of integers ( x,y ) with 1 ≤ x ≤ ( p 1) / 2 and 1 ≤ y ≤ ( q 1) / 2. Note that S contains no pair ( x,y ) with qx = py , so let us partition S into two subsets: S 1 contains all pairs ( x,y ) with qx > py , and S 2 contains all pairs ( x,y ) with qx < py . Note that ( x,y ) ∈ S 1 if and only if 1 ≤ x ≤ ( p 1) / 2 and 1 ≤ y ≤ b qx/p c . So  S 1  = ∑ ( p 1) / 2 x =1 b qx/p c . Similarly,  S 2  = ∑ ( q 1) / 2 y =1 b py/q c . So we have p 1 2 q 1 2 =  S  =  S 1  +  S 2  = ( p 1) / 2 X x =1 b qx/p c + ( q 1) / 2 X y =1 b py/q c , and Theorem 9.6 implies ( p  q )( q  p ) = ( 1) p 1 2 q 1 2 . That proves the first statement of the theorem. The second statement follows immediately. 2 9.3 The Jacobi Symbol Let a,n be integers, where n is positive and odd, so that n = q 1 ··· q k , where the q i are odd primes, not necessarily distinct. Then the Jacobi symbol ( a  n ) is defined as ( a  n ) := ( a  q 1 ) ··· ( a  q k ) , where ( a  q j ) is the Legendre symbol. Note that ( a  1) = 1 for all a ∈ Z . Thus, the Jacobi symbol essentially extends the domain of definition of the Legendre symbol. Note that ( a  n ) ∈ { , ± 1 } . Theorem 9.8 Let m,n be positive, odd integers, an let a,b be integers. Then 1. ( ab  n ) = ( a  n )( b  n ) ; 2. ( a  mn ) = ( a  m )( a  n ) ; 3. a ≡ b (mod n ) imples ( a  n ) = ( b  n ) ; 4. ( 1  n ) = ( 1) ( n 1) / 2 ; 5. (2  n ) = ( 1) ( n 2 1) / 8 ; 6. if gcd( m,n ) = 1 , then ( m  n )( n  m ) = ( 1) m 1 2 n 1 2 ....
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 Spring '13
 MRR
 Math, Algebra, Number Theory

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