Math109Fall2019HW3SolutionsRubric.pdf - Homework 3 Partial...

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Homework 3 Partial Solutions Problems IV 1. Prove that if an integer n is the sum of two squares ( n = a 2 + b 2 for some a, b Z ) then n = 4 q or n = 4 q + 1 or n = 4 q + 2 for some q Z . Deduce that 1234567 cannot be wrtten as the sum of two squares. Proof. Suppose n = a 2 + b 2 for some a, b Z . By exercise 15.5, any perfect square is equal to 4 q or 4 q + 1 for some q Z . This gives three cases for a 2 , b 2 : If a 2 = 4 p and b 2 = 4 q for some p, q Z , then n = a 2 + b 2 = 4 p + 4 q = 4( p + q ) . If a 2 = 4 p and b 2 = 4 q + 1 for some p, q Z , then n = a 2 + b 2 = 4 p + 4 q + 1 = 4( p + q ) + 1 . We obtain a similar result when a 2 = 4 p + 1 and b 2 = 4 q for some p, q Z . If a 2 = 4 p + 1 and b 2 = 4 q + 1 for some p, q Z , then n = a 2 + b 2 = 4 p + 1 + 4 q + 1 = 4( p + q ) + 2 . Therefore, n = 4 q or n = 4 q + 1 or n = 4 q + 2 for some q Z . Since 1234567 = 4(308641) + 3, 1234567 is not the sum of two squares. 2. Let a be an integer. Prove that a 2 is divisble by 5 if and only if a is divisible by 5. Proof. ( ) Suppose a 2 is divisible by 5. We may express a in one of the following ways: If a = 5 q for some q Z , then 5 divides a . If a = 5 q + 1 for some q Z , then a 2
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