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Unformatted text preview: Math 465 Solution Set 3 Ae Ja Yee 1. Prove that if a and b are integers of the form 6 n + 1 or 6 n + 3, then the product ab is also of the form 6 n + 1 or 6 n + 3. Solution. ab = (6 m + 1)(6 n + 1) = 6(6 mn + m + n ) + 1 , ab = (6 m + 1)(6 n + 3) = 6(6 mn + 3 m + n ) + 3 , ab = (6 m + 3)(6 n + 3) = 6(6 mn + 3 m + 3 n + 1) + 3 . 2 2. Prove that there are infinitely many primes of the form 6 n + 5. Solution. Suppose that there are only finitely many primes p = 6 n + 5. Call these primes p 1 ,p 2 ,...,p k with p 1 = 5 ,p 2 = 11. Now consider N = 6 p 2 ··· p k + 5 . If N is prime, then N is not equal to any of the primes p 1 ,p 2 ,...,p k since N > p i for each i . This is a contraction. If N is not prime, then by the unique factorization theorem, we can write N = q 1 ··· q r , where q 1 ,...,q r are primes. The prime 5 does not divide N , for if 5  N , then 5  ( N 5), which is a contradiction....
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 Spring '08
 YEE
 Number Theory, Integers, Prime number, 6m, Fundamental theorem of arithmetic

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