Can we find a positive square integer $k>1$, which satisfies $k\mid 2^{k-1} -1$ ? If yes, what are such $k>1$ values? Here $k = n^2$ and $n$ is some positive integer. If we cannot find such square integer $k>1$, then how to disprove the statement $k\mid 2^{k-1} -1$?
High regards and advanced thanks!
Richard Sieman
Given that $n^2|(2^{n^2-1}-1)$. Now it is obvious that $(2^r-1)|(2^{n^2-1}-1)$, where $r|(n^2-1)$, hence, $2^r-1=n^2$ for some positive integer $r$. Now $r$ cannot be even, (otherwise, $2^{2k}-n^2=1$ for $r=2k$, and factorising yield $2^k-n=1$ and $2^k+n=1$ for which $n=0$, or $2^k-n=-1$ and $2^k+n=-1$ for which $n=1$) so let $r=2k+1$, hence, $2^r-1=n^2$ becomes; $2^{2k+1}-1=n^2$ or $n^2-2(2^k)^2=-1$ which is Pell's equation for which all solutions are given by; $n+(2^k){\sqrt2} \equiv (1+{\sqrt2})^{2t+1}$ for all non-negative integers $t$, $r=(2k+1)|(n^2-1)$, and can be used to narrow your search