Suppose that $2^b-1\mid 2^a+1$. Show that $b = 1$ or $2$. I'm stuck with this one. I would appreciate any idea how to prove this.
 A: Your condition is equivalent to the claim that $2^a\equiv -1 \pmod {2^b-1}$. Look at powers of $2$ modulo $2^b-1$ for different values of $b$, and note that, when $b>2$, none of them are congruent to $-1$. Can you see why?
Does that help?
A: we have $2^{n}-1\mid 2^{2n}-1$ because $$2^{2n}-1=(2^{n}-1)(2^{n}+1)$$ so$$2^{2^b}-1|2^{2^{b+1}}-1$$
which yields that if $a>b$ then $2^{2^b}-1\mid 2^{2^a}-1$. If $2^{2^b}-1\mid 2^{2^a}+1$ then
$$2^{2^b}-1\mid 2=(2^{2^a}+1)-(2^{2^a}-1)$$ which is a contradiction because $2^{2^b}-1$ is an odd number and isn't $1$!
this is an special case of your problem and i guess not to be true in general case!
if $b$ is even then $a$ should be odd because $3\mid 2^{b}-1$ but $2^{a}+1$ is not divisible by $3$.
A: Let $a=qb+r$, where $0\le r<b$. Then, since $2^b-1\mid 2^a+1$ and $2^a\equiv 2^r\pmod{2^b-1}$, we have $2^b-1\mid 2^r+1$.
But we have $0\le r<b$, so this should be impossible in most cases.
Can you continue now?
A: Suppose that $2^b-1 \mid 2^a+1$.
Note that $b$ is the order of $2 \pmod{2^b-1}$. Since $2^{2a} \equiv 1 \pmod{2^b-1}$ we have $b \mid 2a$. 
If $b$ is odd then we get $b \mid a$, so $2^a \equiv 1 \pmod{2^b-1}$. However we also have $2^a \equiv -1 \pmod{2^b-1}$, whence $-1 \equiv 1 \pmod{2^b-1}$, so $b=1$.
Otherwise $b$ is even, so $3 \mid 2^b-1$, so $3 \mid 2^a+1$, so $a$ is odd. Write $b=2c, c \in \mathbb{Z}^+$, so $b \mid 2a$ becomes $c \mid a$. Write $a=kc$. Since $a$ is odd, both $k$ and $c$ are odd. We now get $2^{2c}-1 \mid 2^{kc}+1$.
Thus $(2^c-1)(2^c+1) \mid (2^c+1)(2^{(k-1)c}-2^{(k-2)c}+ \ldots -2^c+1)$
$$2^c-1 \mid (2^{(k-1)c}-2^{(k-2)c}+ \ldots -2^c+1)=\sum_{j=0}^{k-1}{(-1)^j2^{jc}}$$
Now $$\sum_{j=0}^{k-1}{(-1)^j2^{jc}} \equiv \sum_{j=0}^{k-1}{(-1)^j} \equiv 1 \pmod{2^c-1}$$
Therefore $2^c-1 \mid 1$, whence $c=1$, so $b=2$.
In conclusion, $2^b-1 \mid 2^a+1 \Rightarrow b=1, 2$.
A: A bit of help:
Assume $$b\mid a$$ by recurrence of Mersennes: $$2^a-1\equiv 0 \pmod{2^b-1}$$ adding 2, we get that 2 is the remainder for $2^a+1$ . Apply the recurrence for $2^a+1$ form and we get the next few will be $$3, 5, 9, 17, 33,65,129,\ldots$$ in remainders. Anyways you'll find that every second number is divisible by $3$ remainder wise,  and the others have remainder 2 on division by 3 ...
