Is the conjecture true? $3^n - 2^m = 1$ has infinitely many solutions, where n, m are natural numbers.

Is the conjecture true?

$$3^n - 2^m = 1$$ has infinitely many solutions, where n, m are natural numbers.

More generally, $$P^n-Q^m=1$$ has infinitely many solution for n, m ϵ {1,2,3,…} where P is odd, Q is even, P and Q have no common factors.

$$P^n-Q^m=2$$ has infinitely many solution for n, m ϵ {1,2,3,…} where P, Q are both odd, P and Q have no common factors.

This is my first post - I joined this community to look into this.

• Welcome to MSE! Please use the basic tutorial and quick reference guide and also show the work you have done so far. Feb 7, 2021 at 18:49
• See the Catalan conjecture, which was proven in 2002. Feb 7, 2021 at 18:49
• Well, what have you done so far? What solutions have you found so far? If there are infinitely many, there is probably some pattern. So find some! Feb 7, 2021 at 18:49
• Thank you. I'd mark the Catalan conjecture as answering my question. But I don't see how. Feb 7, 2021 at 18:52
• There is only ONE pair of consecutive perfect powers (if we exclude $0$ and $1$) , namely $8$ and $9$. This has been proven in the mean time, for other differences the problem is open (See Pillai's conjecture) Feb 7, 2021 at 18:56

The only natural numbers $$m,n>1$$ and $$P,Q>0$$ satisfying $$P^n-Q^m=1,$$are $$(m,n,P,Q)=(2,3,3,2)$$.