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Unformatted text preview: a 1 , half of the a 2 s make the dierence even and half make the dierence odd. So nally half of the pairs have solutions. p201 E15 For (i) and (iii) you can uniformly prove the following: if r is an odd prime number < q , then p 1( mod r ) cannot happen. Suppose not, i.e., p 1( mod r ). Then by plugging in p = 2 q + 1 we get q 0( mod r ), which for sure doesnt hold, as q is a prime number dierent from r . For (ii) we use the similar argument. The only dierence is that you get 2 q 0( mod 4), therefore q 0( mod 2), which also doesnt hold. p205, E5 The kernel is { [0] , [12] } and for the image, see p200 E12. 1...
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 Spring '08
 BILLERA
 Math, Algebra

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