Note that p 1 2 x j 1 j p 2 1 8 which implies a 1 p 2

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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
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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 ) ∈ { 0 , ± 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 . 59
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Proof. Parts (1)–(3) follow directly from the definition (exercise). For parts (4) and (6), one can easily verify (exercise) that for odd integers n 1 , . . . , n k , k X i =1 ( n i - 1) / 2 ( n 1 · · · n k - 1) / 2 (mod 2) . Part (4) easily follows from this fact, along with part (2) of this theorem and part (1) of Theorem 9.4 (exercise). Part (6) easily follows from this fact, along with parts (1) and (2) of this theorem, and part (5) of Theorem 9.4 (exercise). For part (5), one can easily verify (exercise) that for odd integers n 1 , . . . , n k , X 1 i k ( n 2 i - 1) / 8 ( n 2 1 · · · n 2 k - 1) / 8 (mod 2) . Part (5) easily follows from this fact, along with part (2) of this theorem, and part (4) of Theorem 9.4 (exercise). 2 As we shall see later, this theorem is extremely useful from a computational point of view — with it, one can efficiently compute ( a | n ), without having to know the prime factorization of either a or n . Also, in applying this theorem it is useful to observe that for odd integers m, n , ( - 1) ( n - 1) / 2 = 1 iff n 1 (mod 4); ( - 1) ( n 2 - 1) / 8 = 1 iff n ≡ ± 1 (mod 8); ( - 1) (( m - 1) / 2)(( n - 1) / 2) = 1 iff m 1 (mod 4) or n 1 (mod 4). Finally, we note that if a is a quadratic residue modulo n , then ( a | n ) = 1; however, ( a | n ) = 1 does not imply that a is a quadratic residue modulo n .
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