TITLE: Number theory fun problem
QUESTION [10 upvotes]: Say $a,b > 2 $ are integers. Then we have that $2^a + 1$ is not divisible by $2^b - 1$.
Any thoughts on how to tackle this problem???

REPLY [2 votes]: Hint $\ $ Notice that $\rm\ a \equiv c\pmod{b}\:\Rightarrow\:x^a\equiv x^{c}\pmod{x^b\!-1}\ $ since
$$\rm mod\ x^b\!-1\!:\,\ \color{#C00}{x^b\equiv 1}\:\Rightarrow\:x^a = x^{c+bn} = x^{c} (\color{#C00}{x^b})^n\equiv x^{c} \color{#C00}1^n\equiv x^{c}\quad\ \ $$
Hence for $\rm\,x = 2\,$ and $\rm\: 0 \le c < b,\:$ i.e. $\rm\ c = (a\ mod\ b),\:$ we deduce that
$$\rm mod\ 2^b\!-1\!:\,\ 2^a\! + 1 \equiv 2^{c}\! + 1 < 2^b\!-1\ \  by\ \ \,c< b\ \ and\  hypotheses.$$
Thus we have $\rm\ mod\ 2^b\!-1\!:\,\ 2^a\!+1\not\equiv 0,\ $ hence $\rm\, 2^b\!-1\nmid 2^a\!+1.$  
Remark $\ $ For a more general perspective see here on divisibility sequences.