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On the Wikipedia page, the Beal´s conjecture is stated as:

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?

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  • $\begingroup$ 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. $\endgroup$ – Xarles Jul 16 at 18:24
  • $\begingroup$ @Xarles How large are those "small numbers", generally, "allowed" to be? $\endgroup$ – Grešnik Jul 16 at 18:29
  • $\begingroup$ 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). $\endgroup$ – Xarles Jul 16 at 18:40
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    $\begingroup$ It is not even known, with current technology, that your equation 1) with $x=1$ has finitely many solutions $(B,C,y,z)$. $\endgroup$ – Mike Bennett Jul 17 at 1:25
  • $\begingroup$ @MikeBennett I tried to solve that case, but some difficulties emerged, I think that there are no solutions. $\endgroup$ – Grešnik Jul 17 at 2:45

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