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.
 A: 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 $2,3.$
A: *

*$11^n -1$ is divisible by $10$ and hence never a power of $2.$

*$11^n+1$ is $2\pmod 8$ if $n$ is even and $ 4\pmod 8$ if $n$ is odd. Since it is not divisible by $8,$ it's not equal to any power of $2.$
A: In practice, one virtually never finds an instance of $p^a$ falling adjacent to a power of 2, with the odd exception of cases like $2^3+1 = 3^2$.
The reason for this, is that if some odd $p^a$ is adjacent to a power of 2, then there are some prime factors of $p$ that would divide their own periods.  The only known instance of this is $1093$ and $3511$, but it's not simple powers of these that lie near powers of these numbers.  
For $x^o \pm 1$, where $o$ is odd, one can write this as $(x \pm 1) \sum_{n=0}^{o-1} (\pm x)^a$.  The sumation gives always an odd number, so we might note that if $x^n$ is adjacent a power of 2, then n is itself a power of 2.
For $x^2+1$, one can readily prove that it leaves a remainder of 2 when divided by four, which means the balance of the number is a large odd number.
For $x^2-1$, we might note this is $(x-1)(x+1)$, which has a difference of $2$.  This means that the only solution for powers of $2$ that differ by 2 is 2 and 4.  
This means that the only adjacents to a power of 2, must be of the form of $x\pm 1$, which is hardly a power of $x$.
