Well, its a guess. So please correct me if I'm wrong.
\[(y^9)^4+(10^{25})^4=(x^2)^4\]
This is in \(u^n+v^n=w^n\) form. According to Fermat's last theorem, if n>2 then equation has no integral solution i.e. u,v,n can't be integers simultaneously. The question is in similar form so similar argument can be applied to the question too and it can be said that their is no integral solution of the question.

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TopNewestWell, its a guess. So please correct me if I'm wrong.

\[(y^9)^4+(10^{25})^4=(x^2)^4\] This is in \(u^n+v^n=w^n\) form. According to Fermat's last theorem, if n>2 then equation has no integral solution i.e. u,v,n can't be integers simultaneously. The question is in similar form so similar argument can be applied to the question too and it can be said that their is no integral solution of the question.

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absolutely correct. This is the correct explanation.

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no , it cannot be 101^2 , many of those out of them wont yield integer values of x and y

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Is the answer \(101^2\)?

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