Solution:(Need to confirm )
Need to confirm if this is right
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- Posts:194
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Problem: Find all positive integers $x,y$ such that $p^x-y^p=1$ where $p$ is prime.(Czech Slovakia 1996)
Solution:(Need to confirm )
Solution:(Need to confirm )
Hmm..Hammer...Treat everything as nail
Re: Need to confirm if this is right
I think its correct, however, the solution could be more short, after figuring out, $1+y=p^{x-1}$(Assuming p is odd ofc) , we can say that $\frac{y^p+1}{y+1}=p\Rightarrow 1+y \geq \frac{y^p+1}{y+1} \Rightarrow y^2+2y\geq y^p \Rightarrow y+2\geq y^{p-1}\Rightarrow 2\geq y(y^{p-2}-1)$Asif Hossain wrote: ↑Thu Apr 22, 2021 12:56 pmProblem: Find all positive integers $x,y$ such that $p^x-y^p=1$ where $p$ is prime.(Czech Slovakia 1996)
Solution:(Need to confirm )
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- Posts:194
- Joined:Sat Jan 02, 2021 9:28 pm
Re: Need to confirm if this is right
Nice Solution the idea of bounding didn't come to my mind. mine was rather unnecessarily long
Hmm..Hammer...Treat everything as nail