Is $2^{1093}-2 $ divisible by $1093^2$?

Question

Is $2^{1093}-2 $ divisible by $1093^2$?

I tried to use Euler' theorem with this theorem

$(a,m)=1, a \equiv \alpha \pmod{m}, b \equiv \beta \pmod{\varphi(m)}$ then $a^b \equiv \alpha^\beta $ but not to find solution.

Answer 1

Yes. I did these calculations on my phone:

$1093^2= 1194649.$

$2^4=16$; $2^8=256$; $2^{16}=65536$; $2^{32}=65536^2=4294967296\equiv204141\bmod 1194649$; $2^{64}\equiv204141^2=41673547881\equiv606814\bmod1194649$; $2^{128}\equiv606814^2=368223230596\equiv153273\bmod1194649$; and $2^{256}\equiv153273^2=23492612529\equiv1034593\bmod1194649$.

Therefore $2^{364}\equiv2^42^82^{32}2^{64}2^{256}\equiv16\times256\times204141\times606814\times1034593\equiv 1\bmod1194649$.

Therefore $2^{1093}=(2^{364})^32\equiv2\bmod 1093^2.$

So $1093$ is a Wieferich prime. According to Wikipedia, the only other known Wieferich prime is $3511$.

Answer 2

Alternatively, using MS Excel spreadsheet: $$2^{1093}-2= 2(2^{1092}-1)=2(2^{546}-1)(2^{546}+1)=2(2^{546}-1)(\color{red}{2^{182}+1})(2^{364}-2^{182}+1)\\ ===================================\\ 2^{26}=67,108,864\equiv 75\cdot 1093^2+208,520\equiv 208,520 \pmod{1093^2}\\ 2^{182}\equiv (2^{26})^7\equiv 208,520^7\equiv (208,520^2)^3\cdot 208,520\equiv 145,396^3\cdot 208,520\equiv \\ 145,396^2\cdot 145,396\cdot 208,520\equiv 682,761\cdot 145,396\cdot 208,520\equiv \\ 165,052\cdot 208,520 \equiv -1 \pmod{1093^2=1,194,649}$$