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- #1

- Apr 13, 2013

- 3,720

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...