For $n\in \mathbb N$, $n>1$ prove that $$2^n-1 \neq k^y$$ for all $k,y \in \mathbb N_{\geq 2}.$
Assuming for contradiction that there exists $(k,y)$ such that $2^n-1 = k^y$, I succeeded to prove that the pair does not exist for an even k, and for an even y.
I need to prove that it does also not exist for an odd y.
I need to use in this proof that
$$\frac{x^{2k+1}+1}{x+1} = x^{2k} -x^{2k-1}+\cdots+1.$$
Thank you!
If $y$ is odd (e.g. $y=2z+1$), then:
$$2^n=k^y+1=(k+1)(k^{2z}-k^{2z-1}+\ldots+ 1)$$
This means that the sum in the second brackets on the right has $2z+1$ terms, all being odd, so the whole sum is odd.
This in turn means that $2^n\mid k+1$ as all occurrences of the prime factor $2$ must be present in the first factor $k+1$.
However, as we also have $k+1\mid 2^n$, this means that $k+1=2^n$, i.e. $k=2^n-1=k^y$. So either $k=1$ and so $2^n=2$, i.e. $n=1$ (contradiction), or $k>1$, which implies $y=1$ (contradiction).
