TMUA 2017 · Paper 2 · Question 9
SaveConsider the following attempt to prove this true theorem:
*Theorem:* has no solutions with , and positive integers.
*Attempted proof:* Suppose that there are positive integers , and such that .
I We have .
II Hence .
III It follows that and , since and .
IV Eliminating , we have .
V Multiplying out, we have .
VI Hence so one of and is zero.
But this is a contradiction to the original assumption that all of , and are positive. It follows that the equation has no solutions.
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