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Sean Lawton
Sean Lawton
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how 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$p$ a prime exponent using linear forms in logs?

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

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

how do you bound 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?

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 $p$?

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Kale
Kale
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how do you bound 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?