Divide the equation by the power of 2 that a and b share.
If the right-hand side of the reduced equation = 2, then (a,b,n,m) = (2,2,n,n+1) is a solution of the original equation.
Assume the right-hand side of the reduced equation >=4. Then the 2 terms of the left-hand side can't both be even, since we've factored out common powers of 2, and one can't be odd and the other even, since that would yield a odd sum, so they're both odd. But then their sum = 2mod4, which is impossible.
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Posted by xdog
on 2007-04-09 08:27:10 |