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# 60 chapter 10 computational problems related to

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Chapter 10 Computational Problems Related to Quadratic Residues 10.1 Computing the Jacobi Symbol Suppose we are given an odd, positive integer n , along with an integer a , and we want to compute the Jacobi symbol ( a | n ). Theorem 9.8 suggests the following algorithm: t 1 repeat — loop invariant: n is odd and positive a a rem n if a = 0 if n = 1 return t else return 0 compute a 0 , h such that a = 2 h a 0 and a 0 is odd if h 6≡ 0 (mod 2) and n 6≡ ± 1 (mod 8) then t ← - t if a 0 6≡ 1 (mod 4) and n 6≡ 1 (mod 4) then t ← - t ( a, n ) ( n, a 0 ) forever That this algorithm correctly computes the Jacobi symbol ( a | n ) follows directly from Theo- rem 9.8. Using an analysis similar to that of Euclid’s algorithm, one easily sees that the running time of this algorithm is O ( L ( a ) L ( n )). 10.2 Testing quadratic residuosity 10.2.1 Prime modulus For an odd prime p , we can test if a is a quadratic residue modulo p by either performing the exponentiation a ( p - 1) / 2 rem p or by computing the Legendre symbol ( a | p ). Using a standard repeated squaring algorithm, the former method takes time O ( L ( p ) 3 ), while using the Euclidean- like algorithm of the previous section, the latter method takes time O ( L ( p ) 2 ). So presumably, the latter method is to be preferred. 61
10.2.2 Prime-power modulus For an odd prime p , we know that a is a quadratic residue modulo p e if and only if a is a quadratic residue modulo p . So this case immediately reduces to the previous case. 10.2.3 Composite modulus For odd, composite n , if we know the factorization of n , then we can also determine if a is a quadratic residue modulo n by determining if it is a quadratic residue modulo each prime divisor p of n . However, without knowledge of this factorization (which is in general believed to be hard to compute), there is no efficient algorithm known. We can compute the Jacobi symbol ( a | n ); if this is - 1 or 0, we can conclude that a is not a quadratic residue; otherwise, we cannot conclude much of anything. 10.3 Computing modular square roots 10.3.1 Prime modulus Let p be an odd prime, and suppose that ( a | p ) = 1. Here is one way to compute a square root of a modulo p , assuming we have at hand an integer y such that ( y | p ) = - 1. Let α = [ a mod p ] Z * p and γ = [ y mod p ] Z * p . The above problem is equivalent to finding β Z * p such that β 2 = α . Let us write p - 1 = 2 h m , where m is odd. For any δ Z * p , δ m has order dividing 2 h . Since α 2 h - 1 m = 1, α m has order dividing 2 h - 1 . Since γ 2 h - 1 m = [ - 1 mod p ], γ m has order precisely 2 h . Since there is only one subgroup in Z * p of order 2 h , it follows that γ m generates this subgroup, and that α m = γ mx for 0 x < 2 h and x is even. We can find x by computing the discrete logarithm of α m to the base γ m , using the algorithm in § 8.2.3. Setting κ = γ mx/ 2 , we have κ 2 = α m . We are not quite done, since we now have a square root of α m , and not of α . However, because gcd( m, 2) = 1, we can find integers s, t such that ms + 2 t = 1. In fact, s = 1 and t = -b m/ 2 c do the job. It then follows that ( κ s α t ) 2 = κ 2 s α 2 t = α ms α 2 t = α ms +2 t = α.

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