By Sean Prendiville
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Extra info for Number Theory and Group Theory 2012 with solutions
Sample text
2. Show that the quadratic equation x2 = 1 has four solutions in Z/8Z. Solution: Working mod 8, we have 12 ≡ 1, 32 ≡ 9 ≡ 1, 52 ≡ (−3)2 ≡ 32 ≡ 1 and 72 ≡ (−1)2 ≡ 1. Therefore in Z/8Z the equation x2 = 1 is satisfied when x = 1, 3, 5, 7. 3. In each of the following cases find all the solutions, if there are any, of the given congruence. List the distinct solutions in Z/8Z, not the distinct solutions in Z (of which there could be infinitely many). (a) 6x ≡ 3 (mod 8). (b) 7x ≡ 3 (mod 8). (c) 14x ≡ 6 (mod 8).
List the distinct solutions in Z/8Z, not the distinct solutions in Z (of which there could be infinitely many). (a) 6x ≡ 3 (mod 8). (b) 7x ≡ 3 (mod 8). (c) 14x ≡ 6 (mod 8). Solution: (a) There are no solutions because (6, 8) = 2 which does not divide 3. 1 c University of Bristol [2012] (b) We need integers x and y such that 7x + 8y = 3. A particular solution is x0 = 5, y0 = −4. The general solution is x = 5 + 8k, y = −4 − 7k for k ∈ Z. Therefore the only solution of 7x ≡ 3 (mod 8) is x ≡ 5. e. 7x4 y = 3.
But this follows immediately form Fermat’s little theorem because 560 is divisible by 2, 10 and 16; for instance mod 17 we have (a, 17) = 1 so that a16 ≡ 1 and hence a560 ≡ (a16 )35 ≡ 135 ≡ 1.