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Unformatted text preview: Math 347 C1 HOUR EXAM III 5 August 2009 SOLUTIONS 1. Let a, b, c be integers, relatively prime in pairs, that is, gcd( a, b ) = gcd( a, c ) = gcd( b, c ) = 1. Suppose that a 2 + b 2 = c 2 . Show that a and b must have different parities. SOLUTION Assume that they have the same parity. Since gcd( a, b ) = 1, it follows that both must be odd. Thus let a = 2 r + 1 and b = 2 s + 1 , for some r, s . Then (2 r + 1) 2 + (2 s + 1) 2 = c 2 (4 r 2 + 4 r + 1) + (4 s 2 + 4 s + 1) = c 2 4( r 2 + s 2 + r + s ) + 2 = c 2 . Thus c 2 must be even and so c is even. Let c = 2 u. Then 4( r 2 + s 2 + r + s ) + 2 = c 2 = (2 u ) 2 = 4 u 2 . Therefore, 4 must divide the lefthand side. But then 4  2, a contradiction. Thus a and b have different parities. ANOTHER SOLUTION: The Theorem on Pythagorean Triples states that either a or b must be of the form: 2 rs . Thus either a or b must be even. Since gcd( a, b ) = 1, it follows that they have different parity....
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 Spring '09
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 Math, Pythagorean Theorem, Integers, Mathematical Induction, Natural number

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