I've asked two years ago a post on Mathematics Stack Exchange, were provided two excellent answers. I'm asking on MathOverflow in the hope that some professor can to expand/improve (if it is possible) these results answering my question. The post has the same title and identifier [*3757149*](https://math.stackexchange.com/questions/3757149/on-the-diophantine-equation-xm-1x1-yn-1y1-with-xy-over-integer) on Mathematics Stack Exchange.


I don't know if the following diophantine equation (problem) is in the literature. We consider the diophantine equation $$x^{m-1}(x+1)=y^{n-1}(y+1)\tag{1}$$


over integers $x\geq 2$ and $y\geq 2$ with $x>y$, and over integers $m\geq 2$ and $n\geq 2$. These are four integral variables $x,y,m$ and $n$. The solutions that I know for the problem $(1)$ are two, the solution $(x,y;m,n)=(3,2;2,3)$ and $(98,21;2,3)$.



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> **Question 1.** Do you know if this problem is in the literature? Alternatively, if this problem isn't in the literature can you find more solutions? **Many thanks.**
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If the equation or problem $(1)$ is in the literature please refer it answering this question as a reference request, and I try to search and read the statements for new solutions from the literature. In other case compute more solutions or add upto what uppers limits you got evidence that there aren't more solutions.


I would like to know **what work can be done** with the purpose to know if the problem $(1)$ have finitely many solutions $(x,y;m,n)$.



> 
> **Question 2.** Are there finitely many solutions $(x,y;m,n)$ of stated problem $(1)$? I mean what relevant reasonings or heuristics you can to deduce with the purpose to study if the problem have finitely many solutions. **Many thanks**
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If this second question is in the literature, please refer the literature answering this question as a reference request, and I try to search and read the statements from the literature.



On the diophantine equation $x^{m-1}(x+1)=y^{n-1}(y+1)$ with $x>y$, over integers greater or equal than two
You can make a lot of progress if you're willing to assume a deep conjecutre. The $N$-variable generalization of the $abc$-conjecture (<https://en.wikipedia.org/wiki/N_conjecture>) applied to your equation (with $N=4$) says that if $\gcd(x,y)=1$ and if no subsum of the $4$-term sum
$$ x^n + x^{n-1} - y^m - y^{m-1} $$
vanishes, then
$$ \max\{ x^n, y^m \} \le C\_\epsilon\cdot (xy)^{3+\epsilon}. \quad(\*) $$
Since you're assuming that $x>y\ge2$, this first implies that
$$ x^n \le C\_\epsilon\cdot x^{6+2\epsilon}, $$
so there are only finitely many solutions with $n\ge7$. If we assume that $n\le6$, then
$$ x^6 \le x^n \le x^n + x^{n-1} = y^m + y^{m-1} \le 2y^m, $$
so $x\le (2y^m)^{1/6}$. Then $(\*)$ gives
$$ y^m \le C\_\epsilon (xy)^{3+\epsilon} \le C\_\epsilon ((2y^m)^{1/6})^{3+\epsilon})y^{3+\epsilon} \le C'\_\epsilon y^{(m/2+3)(1+\epsilon)}. $$
This shows that there are only finitely many solutions with $m>m/2+3$, so only finitely many solutions with $m>6$.


So now you're reduced to a finite number of exponents, and since you've assumed that $x>y$, you need to handle exponents satisfying $2\le n<m\le 6$. Presumably for each choice of $(m,n)$, there are only finitely many solutions. In any case, since you want to avoid $x=y$, you're looking at integral points on the curve
$$ \frac{x^n + x^{n-1} - y^m - y^{m-1}}{x-y}=0. $$
For each $(m,n)$, there will only be infinitely many integer solutions if the polynomials has a linear or quadratic factor.


Finally, it's possible that the cases where some subsum vanishes can be treated directly, but in any case, the usual $abc$-conjecture will handle the three term vanishing sums, and the two term vanishing sums are easy.


The assumptions in this analysis include the assumption that $\gcd(x,y)=1$. If you don't want that assumption, you can write $x=ZX$ and $y=ZY$ with $Z=\gcd(x,y)$, so your equation becomes
$$ ZX^n + X^{n-1} = Z^{m-n+1} Y^m + Z^{m-n} Y^{m-1}. $$
Now you can try applying the $4$-variable $abc$ conjecture to this equation.


