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Minor edits.
Sean Lawton
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How do you bound the exponent of x^2+1=y^p?

How do you bound the exponent of $x^2+1=y^p$ for $p$ a prime exponent using linear forms in logs?

So far I have $(x-i)(x+i)=y^p$ which are coprime and hence $x+i=(a+ib)^p$.

Now how do I get a linear form in logs so that I can find an upper bound on $p$?

Kale
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