# sol1 - MATH 115B SOLUTION SET I APRIL 17, 2007 (1) Prove...

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MATH 115B SOLUTION SET I APRIL 17, 2007 (1) Prove that the quadratic congruence 6 x 2 + 5 x + 1 0 mod p has a solution modulo every prime p , even though the equation 6 x 2 + 5 x + 1 = 0 has no solution in the integers. Solution: First observe that, for p = 2, the congruence 6 x 2 + 5 x + 1 0 (mod 2) has a solution x 1 (mod 2). Also, when p = 3, the congruence 6 x 2 + 5 x + 1 0 (mod 3) has a solution x 1 (mod 3). Suppose therefore that p is a prime with p > 3. Then the congruence 6 x 2 + 5 x + 1 0 (mod p ) may be expressed as y 2 1 (mod p ), where y 12 x + 5 (mod p ). The solution y 1 (mod p ) leads to the linear congruence 12 x ≡ - 4 (mod p ). Since (12 , p ) = 1, this congruence has a solution. It therefore follows that the congruence 6 x 2 +5 x +1 0 (mod p ) also has a solution. (2) Show that 3 is a quadratic residue modulo 23, but is a non-residue modulo 31. Solution: In the case of p = 23, Euler’s criterion yields 3 (23 - 1) / 2 = 3 11 1 (mod 23) , and so 3 is a quadratic residue modulo 23. For p = 31, we have 3 (31 - 1) / 2 = 3 15 ≡ - 1 (mod 31) , and so 3 is a quadratic non-residue modulo 31. (3) Given that a is a quadratic residue modulo the odd prime p , prove the following: (a) a is not a primitive root of p . (b) The integer p - a is a quadratic residue or non-residue modulo p according as p 1 (mod 4) or p 3 (mod 4). (c) If

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sol1 - MATH 115B SOLUTION SET I APRIL 17, 2007 (1) Prove...

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