If m is odd and n a natural number (2ᵐ-1,2ⁿ+1)=1.

In the following i denote with (a,b) the greatest common divisor between two integers a and b.

Let p now be a possible prime divisor of A=2ᵐ-1 and B=2ⁿ+1.

Let [2]_p denote the class of 2 modulo p.

Let us note that since (2,p)=1 then [2]_p is an element of the group of invertibles of Zp, which I denote with U(Zp) (explicitly U(Zp)=({[a]_p : (a,p)=1} ,*) where * is the natural product between equivalence classes).

We observe that, since p|A:

•)[2]ᵐ_p=[2ᵐ]_p=[1]_p

and since p|B,

[2]ⁿ_p=[2ⁿ]_p=[-1]_p that is:

••)[2]²ⁿ=[1]_p

therefore from •) and ••) we have that the order o([2]_p) of [2]_p in U(Zp) must be a divisor of m and of 2n.

So since m is odd we have that o([2]_p) divides (m,2n)=(m,n). Therefore:

[2]ⁿ_p=[1]_p

which leads, taking into account ••) , to the equality:

[1]_p=[-1]_p

that is, [2]_p=[0]_p, which is a contradiction since p is an odd prime.

Consequently, A and B have no prime divisors in common, that is, (A,B)=1.