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TMUA 2017 · Paper 2 · Question 9

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Consider 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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