TITLE: Proof $2^{1092}-1$ is divisible by $1093^2$?
QUESTION [5 upvotes]: I would like a proof that $2^{1092}-1$ is divisible by $1093^2$. I can prove it is divisible by $1093$ by using Fermat ($1093$ is a prime) or Euler. However I'm pretty sure we are going to have to look at more things to crack this one. I also tried to write $1093$ in binary $10001000101$ and see if it has multiples made entirely of ones, but that didn't work either. Thank you very much.
Regards.

REPLY [5 votes]: Thanks to the help of Daniel Fischer and others in the chat I now have a viable solution by a procedure known as exponentiation by squaring:
$2^{17}= 131072\leadsto (2^{17})^2= 2^{34} \equiv 816564\equiv -378085 \leadsto$
$(2^{34})^2=2^{68}\equiv 1042817\equiv -151832 \leadsto$
$(2^{68})^2=2^{136}\equiv 1009120\equiv -185529 \leadsto$
$(2^{136})^2=2^{272}\equiv 782853 \leadsto$
$2^{272}\cdot 2=2^{273} \equiv 371057 \leadsto$
$(2^{273})^2=2^{546}\equiv 1194648\equiv-1 \leadsto$
$(2^{546})^2=2^{1092}\equiv 1 \bmod 1093^2$. As desired.