# Microsoft_Word_-_module_IV_quadratic_Resdues.pdf - 1 Module...

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1 Module IV Quadratic Residues: The Legendre and Jacobi Symbols The apparent difficulty of determining quadratic residues (the Quadratic Residuosity Problem) is the basis for believing the Goldwassor-Micali probabilistic public-key encryption scheme to be secure. For this reason we study quadratic residues. n 2 n Definition 1. Let a Z . Then a is a quadratic residue modulo n (or a square modulo n) if and only if the equation x = a has a solution in Z . otherwise it is a quadratic non-residue modulo n. Th * * n n n n n 2 e set of quadratic residues modulo n is denoted by Q ; the non-residues by Q (so that Z = Q Q ). More generally if gcd(a, n) = 1, a is a quadratic residue if and only if x a (mod n) has a solut * n n 2 n 2 ion. Proposition 1. Consider a Z and b a(mod n). Then b is a quadratic residue if and only if a Q . Futhermore, y is a solution of y b(mod n) if and only if y x (mod n) for some x Z such that x * * a (mod n). 2 2 n 2 Proof Suppose b a (mod n) and x a (mod n) where x Z . Then x b (mod n) as well so ˆ ˆ that b is a quadratic residue mod n. Next suppose that x such that x b (mod n). Now gcd (b, n) = gcd (a, n) = * n 2 2 n 2 ˆ 1 so gcd (x, n) = 1 as well. Let x denote the unique element of Z such that ˆ ˆ x x (mod n). Then x x b a (mod n) and a Q follows. The above argument also shows that if y is a solution of y * 2 n 2 2 2 b (mod n) then y x (mod n) where x Z and satisfies x a (mod n). Of course if y x (mod n) and x a (mod n) then y x a b (mod n) thereby concluding the proof of the proposition. * p n i p p p 2k p Proposition 2. Let p be an odd prime and be a generator of Z . Then a Q if and only if a (in Z ) where i p-1 and i is even. p-1 Thus Q = = Q . 2 Proof ( ): If a = (in Z ) where 2k p-1 then x α α α * k p 2 n p j 2j p = (in Z ) is a solution. ( ): If a Q then x such that x = a in Z But x = for some 0 j p-2. Hence a = in Z Write 2 j = q(p-1) + r such that 0 r p-2. Then r is even α α α 2j r p and a = = in Z . α α 17 Example 1. Consider Z ; it has 8 generators one of which is 3. We list the powers of = 3 below: α *

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2 i i α 0 1 1 3 2 9 3 10 4 13 5 5 6 15 7 11 8 16 9 14 10 8 11 7 12 4 13 12 14 2 15 6 Thus { } { } 17 17 Q = 1, 9, 13, 15, 16, 8, 4, 2 and Q = 3, 10, 5, 11, 14, 7, 12, 6 2 n p Observation 1. If x is a solution of x = a in Z the so is - x (because in a ring with identity 1, (-1)(-1)=1). Proposition 3. If p is an odd prime and a Q then a has exactly two square root p 2 2 p s in Z (i.e. solutions of x = a). Proof Recall from Lemma 2 of Module III that x - a = 0 can have at most two solutions. Also x - x in Z . * p Our next result is a slight modification of Proposition 2. It identifies the elements of Q in a more elementary way. We illustrate it first with the aid of the previous example. { } 2 2 2 2 2 2 2 2 17 Example 1. revisited: Realize that Q = 1=1 ,4 = 2 ,9 = 3 ,16 = 4 ,8 = 5 (mod 17),2 = 6 (mod 17),15 = 7 (mod 17), 13 = 8 (mod 17) .
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