# Number TheoryHow can I continue,in order to show that 2^b-1 does not divide 2^a+1?

#### evinda

##### Well-known member
MHB Site Helper
Hi!!! I am looking at the following exercise:
If $a,b \geq 3$,prove that $2^b-1$ does not divide $2^a+1$.
That's what I have tried so far:
We suppoe that $2^b-1|2^a+1$.
We know that $2^b-1|2^b-1$.
So,we get that $2^b-1|2^a+2^b$.
But how can I continue? Do,I have to show that $(2^b-1,2^a+2^b)=1$ ?
I have tried to do this,like that: Let $(2^b-1,2^a+2^b)=d>1$,so $d$ has a prime divisor,$p$.
$p|d,d|2^b-1,d|2^a+2^b \Rightarrow p|2^b-1,p|2^a+2^b$ ,but I don't know how I could continue... #### Prove It

##### Well-known member
MHB Math Helper
Hi!!! I am looking at the following exercise:
If $a,b \geq 3$,prove that $2^b-1$ does not divide $2^a+1$.
That's what I have tried so far:
We suppoe that $2^b-1|2^a+1$.
We know that $2^b-1|2^b-1$.
So,we get that $2^b-1|2^a+2^b$.
But how can I continue? Do,I have to show that $(2^b-1,2^a+2^b)=1$ ?
I have tried to do this,like that: Let $(2^b-1,2^a+2^b)=d>1$,so $d$ has a prime divisor,$p$.
$p|d,d|2^b-1,d|2^a+2^b \Rightarrow p|2^b-1,p|2^a+2^b$ ,but I don't know how I could continue... Your statement is false, if a = b then $2^b - 1$ DOES divide $2^a - 1$...

##### Well-known member
Hi!!! I am looking at the following exercise:
If $a,b \geq 3$,prove that $2^b-1$ does not divide $2^a+1$.
That's what I have tried so far:
We suppoe that $2^b-1|2^a+1$.
We know that $2^b-1|2^b-1$.
So,we get that $2^b-1|2^a+2^b$.
But how can I continue? Do,I have to show that $(2^b-1,2^a+2^b)=1$ ?
I have tried to do this,like that: Let $(2^b-1,2^a+2^b)=d>1$,so $d$ has a prime divisor,$p$.
$p|d,d|2^b-1,d|2^a+2^b \Rightarrow p|2^b-1,p|2^a+2^b$ ,but I don't know how I could continue... There are 2 cases
case 1
b is a factor of a say a = mb

then $2^b-1$ is a factor of $2^{mb} - 1$ so $2^b-1$ is not a factor of $2^a + 1$ as remainder = 2
case 2
b is not a factor of a so a = mb + c where c < b

$2^a+ 1 = 2^{mb+c} + 1$
= $2^c ( 2^{mb}- 1) + 2^c + 1$
now $2^b$ devides $2^c ( 2^{mb}- 1)$ but as c < b
$2^c + 1 < 2^b-1 as 2^c + 2 < 2^{c+1}$

so it does not devide