Divisibility by large powers Show that $2^{111} + 1$ divides $2^{555} + 1$ but does not divide $2^{444} + 1$.
My try:
$(2^{111}+1)^5 = 2^{555}+5*2^{444}+10*2^{333}+10*2^{222}+5*2^{111}+1$
$2^{555} + 1$ = $(2^{111}+1)^5$ - $(5*2^{444}+10*2^{333}+10*2^{222}+5*2^{111})$
= $(2^{111}+1)^5$ - ($5*2^{333}(2^{111}+1)+5*2^{222}(2^{111}+1)+5*2^{111}(2^{111}+1))$
Each term is divisible by $(2^{111} + 1)$ and hence divisible by $2^{111} + 1$.
Such a nice distribution is not possible for $2^{444} + 1$
I could answer the first part and the second part of indivisibility is not as clean as I would like.  Could someone let me know how I could prove the second part.

Note that $2^{111}\equiv -1\pmod{2^{111}+1}$. Thus $2^{555}\equiv (-1)^5=-1\pmod{2^{111}+1}$.
Similarly, $2^{444}\equiv (-1)^4=1\pmod{2^{111}+1}$, and therefore $2^{111}+1$ divides $2^{444}-1$. It follows that $2^{111}+1$ cannot divide $2^{444}+1$. 
