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

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