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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Shubham Srivastava
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3 years, 4 months ago

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absolutely correct. This is the correct explanation.
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Subharthi Chowdhuri
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3 years, 4 months ago

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no , it cannot be 101^2 , many of those out of them wont yield integer values of x and y
–
Subharthi Chowdhuri
·
3 years, 4 months ago

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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. – Shubham Srivastava · 3 years, 4 months ago

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absolutely correct. This is the correct explanation. – Subharthi Chowdhuri · 3 years, 4 months ago

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no , it cannot be 101^2 , many of those out of them wont yield integer values of x and y – Subharthi Chowdhuri · 3 years, 4 months ago

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Is the answer \(101^2\)? – Vikram Waradpande · 3 years, 4 months ago

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