When does $2^{2m}+2^m+1$ divide $10^{2m}+10^m+1$? For what values of $m\in \Bbb N$ is it true that $2^{2m}+2^m+1$ divides $10^{2m}+10^m+1$?
I have it false for $m=1,3 $ and true for $m=2,4$
This arises from my thoughts on the question How to find all binary numbers in base $10$ s.t. that its divisible by its own numerical value in base $2$?

partial answer:  using my comment we get:
 $\begin{align} \\
2^{2m}+2^m+1 &\mid 2^{2m}5^{2m}+2^m5^m+1 \\
&\implies 2^{2m}+2^m+1 \mid 2^{2m}(5^{2m}-1)+2^m(5^m-1) \\
&\implies 2^{2m}+2^m+1 \mid 2^m\big({5^{2m}-1\over 4}\big)+\big({5^m-1\over 4}\big) \\
&\implies 2^m(2^m+1)+1 \mid 2^m\big({5^{2m}-1\over 4}\big)+\big({5^m-1\over 4}\big)
\end{align}$ 
hope this helps simplify things. edit: the fractions are the base 5 repunits. 
LATE Edit:  if M is divisible by totient function values you can also use Euler's theorem to prove that both are 3 mod what ever numbers give those totients. then it's about when two numbers on arithmetic progressions divide each other.
