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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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 Fall '11
 Mitchelle

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