<s>[INST] Number theory: Odd number and powers of 2 Is is true that for any odd natural number $x > 2$, there exists a positive natural number $y$, such that $x^y = 2^n+1$ or $x^y=2^n-1$ where $n$ is a also natural $> 0$. This cannot be solved by simple group theory methods, since we demand that $x^y$ be exactly $2^n+1$ or $2^n-1$ and not only modulu $2^n$.
Thanks.
 [/INST] This is to show without the Catalan conjecture/theorem that $11^y\pm 1$ cannot be a power of $2$. Begin with the polynomial $p=x^2-x-1$ whose value at $x=4$ is $11.$ Note that the last two terms in the expansion of $p^y$ are $yx+1$ if $y$ is even, and $-yx-1$ if $y$ is odd. This means that if $y$ is even then $p^y+1$ is $2$ mod $4$ [so not a power of $2$] and also that if $y$ is odd then $p^y-1$ is also $2$ mod $4$. So we may assume either $y$ is even and we're looking at $p^y-1$ or else $y$ is odd and we look at $p^y+1,$ in both cases wanting to rule out getting a power of $2$.
Now if $y$ is even, we have for $y=2$ that $p^2-1=(x-2)(x-1)x(x+1)$, and since for larger even $y$ we have $p^y-1$ divisible by $p^2-1$ we have in all these cases the four divisors $2,3,4,5$ (recalling here $x=4$), and so $p^y-1$ is not a power of $2$. 
On the other hand if $y$ is odd we have for $y=1$ that $p^1+1=x(x-1)$, and for larger odd $y$ we have $p^y-1$ divisible by $p-1$, so in these cases $p^y+1$ is divisible by $3,4$ so cannot be a power of $2$.
So there is an elementary way to show that $11^y\pm 1$ is not a power of $2$, without the need of a deep theorem like Catalan's.
Edit: I realized after entering this that it can be done without the polynomial. For $y$ even, $11^y+1$ is 2 mod 4 (because $11$ is odd its even powers are $1$ mod 4) Then since $11^2-1$ is divisible by $2,3,5$ we have for larger even integers that $11^y-1$ is divisible by $11^2-1$, so also by $2,3,5.$ For $y$ odd we have that $11^y-1$ is 2 mod 4 while $11^1+1$ is divisible by $2,3$ and then for larger $y$ we have $11^y+1$ is divisible by $11^1+1$ and so also by $2,3.$
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