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Suppose that $(k,l,m) \in \mathbb{N_0}^3$.

If $(k,l,m)=(0,0,0)$ then for $n=1,2$ there is an infinite number of solutions and, by the theorem of of Wiles there are no solutions when $n \geq 3$.

Is it known can there be an infinite number of solutions of $a^{n+k}+b^{n+l}=c^{n+m}$ if $k,l,m$ are not all equal and $n \geq 3$?

What is known of when there is a finite non-zero number of solutions?

Can someone summarize all that is known about this generalized Fermat problem?

We see that a mapping $m: (k,l,m) \to a^{n+k}+b^{n+l}=c^{n+m}$ is a mapping that sends some element of $\mathbb{N_0}^3$ to a family of equations over the integers and because every such equation has its own set of solutions for every $n \in \mathbb N$ we can define some mapping $s$ that sends some element of $\mathbb{N_0}^3$ to a set of all solutions.

A question can then be phrased of when is $s(k,l,m)$ an empty set and when it is not an empty set.

We could probably somehow lower "dimensionality" of this problem by choosing some $(k,l,m)$ and fixing either $a$ or $b$ or $c$ but I am not sure how general results could be obtained in such a way.

Also, I do not know are techniques of Wiles suited for this more general problem, and, if they are not, I do not know why they are not?

It would be nice if someone could explain in non-strict technical sense all that is known about this generalized Fermat problem.

If this is off-topic and written in non-professional way then please vote to close this question of mine.

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    $\begingroup$ It is common to insist in this sort of problem that $a,b,c$ be relatively prime (as GH from MO does in his answer). Without that insistence, you can make as many solutions as you want. E.g., from $3^3+2^4=43$, we can get $a^3+b^4=c^5$ with $a=3\times43^8$, $b=2\times43^6$, $c=43^5$. $\endgroup$
    – Gerry Myerson
    Commented Apr 5, 2018 at 3:55

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Darmon and Granville proved, using Faltings' Theorem, that your equation has finitely many primitive integer solutions for any fixed exponents which are at least $3$. In fact their result is more general. For several concrete exponents beyond Fermat's Last Theorem, the full set of primitive integer solutions is also known, see e.g. the papers of Siksek-Stoll and Anni-Siksek.

You can find more information in the Wikipedia article on Beal's conjecture and the references therein. See also this survey, especially Section 4.5.

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  • $\begingroup$ I note that OP never specifies primitivity. OP might be interested in knowing that if $a,b,c$ are permitted to have common factors, then there are trivial ways to produce lots of solutions. $\endgroup$
    – Gerry Myerson
    Commented Apr 4, 2018 at 23:31
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    $\begingroup$ @GerryMyerson: I agree. Please mention this comment under the original post as well (so that the OP is notified about it). $\endgroup$
    – GH from MO
    Commented Apr 5, 2018 at 1:23

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