The question is easy to verify but hard to prove. Can anyone suggest some idea? I have one approach using Pell's equation, but I don't quite get it .
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1What is the difference between proving and verifying? – The Count Mar 21 '17 at 02:05
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1It seems this follows from Catalan's conjecture (now Mihăilescu's theorem). – user397701 Mar 21 '17 at 02:10
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First assume $k>0$. Suppose that $$2^k+1=\Bigl(\frac xy\Bigr)^2$$ where $x,y$ are positive integers which, without loss of generality, have no common factor. We have $$y^2(2^k+1)=x^2\ ;$$ this means that any prime factor of $y$ is also a factor of $x$, and hence $y$ cannot have any prime factors. That is, $y=1$. Therefore $$2^k=x^2-1=(x-1)(x+1)$$ and so $x-1$ and $x+1$ are both powers of $2$. The only powers of $2$ differing by $2$ are $2^1$ and $2^2$, so $x=3$ and $k=3$.
If $k<0$, say $k=-m$, we have $$y^2(1+2^m)=x^22^m\ .$$ As before, $y$ has no odd prime factor, and by unique factorisation $y^2=2^m$, so $m$ is even. But then we have $1+2^m=x^2$ and the above argument gives $m=3$. So there is no solution for $k<0$.
Trivially, $k=0$ is not a solution.
David
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