TITLE: Is $2^{1093}-2$ a multiple of $1093^2$?
QUESTION [8 upvotes]: I can't solve this problem; it may be easy though.
Is the number $2^{1093} -2$ a multiple of $1093^2$?
I do appreciate any kind of help.

REPLY [3 votes]: (Note:  This answer essentially duplicates the one given by Bill Dubuque.  This is due to the merger of a question asked Mar 2 '15 with a duplicate from Jan 9 '12.)
In a 1983 paper in The Mathematical Intelligencer titled simply "1093," Paolo Ribenboim gives the following proof, with credit to Landau:

Let $p=1093$.
$$3^7=2187=2p+1$$ so squaring
$$3^{14}\equiv4p+1\pmod {p^2}\qquad(*)$$ On the other hand,
  $$2^{14}=16384=15p-11$$ so $$2^{28}\equiv-330p+121\pmod{p^2}$$ hence
  $$3^2\times2^{28}\equiv-1876p-4\pmod{p^2}$$ and dividing by $4$
  $$3^2\times2^{26}\equiv-469p-1\pmod{p^2}$$ Raising now to the
  $7^\text{th}$ power:
$$3^{14}\times2^{182}\equiv-4p-1\pmod{p^2}$$
and taking $(*)$ into account
$$2^{182}\equiv-1\pmod{p^2}$$
Finally, raising to the $6^\text{th}$ power,
$$2^{1092}\equiv1\pmod{p^2}$$

One point I found necessary to think about is $$(1+469p)^7\equiv1+7\cdot469p=1+3283p\equiv1+4p\pmod {p^2}$$
Aside from that, everything is straightforward calculation with small numbers.  How Landau came up with this approach is the real marvel here.
It's perhaps worth noting that this proof does not rely on $1093$ being prime.  It only uses the fact that it's not divisible by either $2$ or $3$ (in the steps where you divide by $4$ and cancel the $3^{14}$.)