exam3takehome - a is even while b and c are odd, say a = 2...

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This is an extra credit problem for the third exam, given Wednesday 3/31/10. This is due by Monday, April 12. It will be worth up to 6 points, to be added to the exam grade. You must include the following signed statement with your answer: On my honor as a University of Florida student, I have not received help from or otherwised discussed this problem with anyone except my instructor. Below I give a sketch of a proof of a theorem. Your assignment is to (i) rewrite this sketch as a proof, in the class style, using either strong induction or the well ordering principle, and (ii) answer the question at the end. Theorem. There are no numbers a,b,c,m Z + such that a 2 + b 2 + c 2 = 2 mabc. (1) Sketch of proof. I claim that all of a , b , and c are even. We can see this by looking at cases. First, if none of them are even, or exactly two of them are even, then the left hand side is odd, while the right hand side is even. The case that exactly one is even is a little harder: We can suppose by symmetry that
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Unformatted text preview: a is even while b and c are odd, say a = 2 a , b = 2 b + 1 and c = 2 c + 1. Then equation (1) becomes 4( a ) 2 + (2 b + 1) 2 + (2 c + 1) 2 = 4 ma bc . Thus the right side is divisible by 4, but on multiplying out the left side we see that it is not. Now since all of a,b,c are even, we can write a = 2 a , b = 2 b and c = 2 c . Then equation (1) becomes (2 a ) 2 + (2 b ) 2 + (2 c ) 2 = 2 m (2 a )(2 b )(2 c ) , or 4( a ) 2 + 4( b ) 2 + 4( c ) 2 = 16 ma b c , which becomes ( a ) 2 + ( b ) 2 + ( c ) 2 = 2(2 m ) a b c , which is a smaller solution to equation (1). Question: Equation (1) does have the trivial solution a = b = c = 0, which is ruled out in the theorem by the requirement that they be members of Z + . How does the proof break down if one attempts to apply it to this trivial solution? 1...
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