# Two equations and a question related to a well-known conjecture from number theory

If $$A^x+B^y=C^z$$, where $$A,B,C,x,y,z$$ are positive integers with $$x,y,z>2$$, then $$A$$,$$B$$, and $$C$$ have a common prime factor.

I think that it is within reach (possibly, this is already solved somewhere) to rephrase the conjecture so as to have $$A,B,C,x,y,z >2$$ as the assumption.

To clarify, the case when $$A=1$$ or $$B=1$$ is already solved and some of us call it Mihăilescu's theorem.

The case $$C=1$$ clearly does not have solutions.

So, the possible rephrasal needs for these cases to be considered and solved:

1) $$2^x+B^y=C^z$$

2) $$A^x+2^y=C^z$$

3) $$A^x+B^y=2^z$$

Of course, 1) and 2) can be considered as exactly the same cases, without any loss of generality.

So I am interested slightly in finding all of the solutions of two not extremely general Diophantine equations, the first task being:

Find all solutions of the equation $$2^x+B^y=C^z$$ where $$B,C,x,y,z \in \mathbb N$$ and $$x,y,z>2$$

and the second one being:

Find all solutions of the equation $$A^x+B^y=2^z$$ where $$A,B,x,y,z \in \mathbb N$$ and $$x,y,z>2$$

I know about Fermat–Catalan conjecture but this question is different, although much related.

So I would like to know are these two tasks somewhere solved and, if they are not, what is, approximately, the current progress made so far on them?

• The cases already solved are when you fix (some of) the exponents to some "small numbers". The problem then reduces to find all rational points of some algebraic curves. – Xarles Jul 16 at 18:24
• @Xarles How large are those "small numbers", generally, "allowed" to be? – Grešnik Jul 16 at 18:29
• You can see in wikipedia page: between 2 and 5, mainly. Other cases are may be known. I don't know if someone has considered fixing one of the basis to 2 (but I am sure someone did, not if they succeeded proving something). – Xarles Jul 16 at 18:40
• It is not even known, with current technology, that your equation 1) with $x=1$ has finitely many solutions $(B,C,y,z)$. – Mike Bennett Jul 17 at 1:25
• @MikeBennett I tried to solve that case, but some difficulties emerged, I think that there are no solutions. – Grešnik Jul 17 at 2:45