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sol6 - Math 465 Solution Set 6 1 Evaluate the following...

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Math 465 Solution Set 6 Ae Ja Yee 1. Evaluate the following Legendre symbols. (Note that 127 is prime.) (a) ( 2 127 ) (b) ( - 1 127 ) (c) ( 5 127 ) (d) ( 15 127 ) Solution. (a) 127 ≡ - 1 (mod 8) , 2 127 = 1 (b) 127 3 (mod 4) , - 1 127 = - 1 (c) 5 127 = ( - 1) 2 · 63 127 5 = 2 5 = - 1 (d) 15 127 = 3 127 5 127 = ( - 1) 1 · 63 127 3 ( - 1) 2 · 63 127 5 = 1 3 = 1 2. How many solutions do the following congruences have? (a) x 2 2 (mod 61) (b) x 2 ≡ - 2 (mod 59) (c) x 2 + 5 x + 1 0 (mod 13) Solution. (a) Since 61 5 (mod 8), 2 61 = - 1 , so no solution exists. (b) Since 59 3 (mod 4) and 59 3 (mod 8) - 2 61 = - 1 61 2 61 = ( - 1)( - 1) = 1 , so there exist solutions. If x 0 is a solution to the congruence, then so is - x 0 , which is not congruent to x 0 , so there are two distinct solutions. 1

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(c) x 2 + 5 x + 1 x 2 - 8 x + 1 ( x - 4) 2 - 2 0 (mod 13) , which is equivalent to ( x - 4) 2 2 (mod 13). Since 13 5 (mod 8), ( 2 13 ) = - 1. This shows that no solution exists. 3. Determine which of the following congruences are solvable. (a) x 2 3 (mod 11 2 · 23 2 ) (b) x 2 9 (mod 2 3 · 3 · 5 2 ) Solution. (a) 3 11 = ( - 1) (3 - 1) / 2 · (11 - 1) / 2 11 3 = - 2 3 = 1 , 3 23 = ( - 1) (3 - 1) / 2 · (23 - 1) / 2 23 3 = - 2 3 = 1 Therefore, the congruence is solvable. (b) The original congruence is equivalent to x 2 9 (mod 2 3 · 5 2 ) x 2 9 (mod 3) Since x 2
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