If $a$ and $b>2$ are any positive integers, then prove that $2^{a}+1$ is not divisible by $2^{b}-1$
Here's my attempt: If $2^{b}-1\mid 2^{a}+1$ then it is obvious that $a>b$. Now ,By division algorithm $$a=qb+r$$
We know that $$2^{b}\equiv 1\pmod{2^{b}-1}$$
Now raising to the power $q$ and then multiplying by $2^r$ we have, $$2^{bq+r}\equiv 2^{r}\pmod{ 2^{b}-1}$$ But $$bq+r=a$$ Therefore, $$2^{a}\equiv 2^{r}\pmod{ 2^{b}-1}$$ Adding one on both the sides $$2^{a}+1\equiv 2^{r}+1\pmod{ 2^{b}-1}$$
As $r<b$, $$2^{r}+1<2^{b}-1$$
Therefore, $$2^{r}+1\not\equiv 0 \pmod{ 2^{b}-1}$$ Therefore, $2^{a}+1$ is not divisible by $2^{b}-1$ .
Can you show me where I went wrong?