Let p be a prime and assume that p ≡ 3 (mod 4). Prove that, if a is a quadratic residue modulo p, then the two square roots of a are ±a^((p+1)/4).
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This is because if -1 were a quadratic residue, then there would exist an integer x such that x^2 ≡ -1 (mod p), which would imply that x^(p-1) ≡ (-1)^((p-1)/2) (mod p) by Fermat's Little Theorem. But since p ≡ 3 (mod 4), we have (p-1)/2 is odd, so (-1)^((p-1)/2) = Show more…
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Let p be a prime satisfying p ≡ 3 (mod 4) and suppose that a is a quadratic residue modulo p. Show that x = a^((p+1)/4) is a solution to the congruence x^2 ≡ a (mod p).
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Quadratic residues. An integer a is called a quadratic residue modulo p if a ≢ 0 (mod p) and there is an integer b such that b^2 ≡ a (mod p) (in other words, a has a square root in Z/pZ). Let p ≡ 3 (mod 4) be an odd prime. (a) Suppose that a is a quadratic residue modulo p. Prove that the only two roots of x^2 - a in Z/pZ are a^{(p+1)/4}. (b) Show that (still under the assumption p ≡ 3 (mod 4)) -1 is not a quadratic residue modulo p.
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