# On the diophantine equation $x^{m-1}(x+1)=y^{n-1}(y+1)$ with $x>y$, over integers greater or equal than two

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 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 $$(98,21;2,3)$$.

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.

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

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.

• When I can, I am going to accept an answer in the site Mathematics Stack Exchange. So the research of this problem will remain in that MathOverflow. Sep 9 at 17:45
• Why not $\ x^m\cdot(x+1)\ =\ y^n\cdot(y+1)\$ over natural $\ x\ y\ m\ n\,\ \ (x>y)$ ? (natural means positive integers). Sep 9 at 19:08
• Many thanks for share this @WlodAA my problem is that I have no internet at home and now were posted answers below thus I can not update the post. Feel free to study you reduction or post yours thoughts as an answer. Many thanks again (I'm not a professional mathematician and I don't realize some details of this kind). Sep 10 at 11:58

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 $$2\le n. Presumably for each choice of $$(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.

• In my answer if we set $m=n$, we get many genus zero curves for $m=n<30$.
– joro
Sep 10 at 6:58
• Many thanks I'm familiarized with this kind of reasonings in the context of an application of abc conjecture for the asymptotic Fermat's Last Theorem; and Mason's theorem. When I can I study your excellent answer. Sep 10 at 12:04
• @joro I don't doubt that one gets lots of genus zero curves. For each of those, there are fairly easy criteria for whether there can be infinitely many integer points, based on the number of points "at infinity" on the completion of the affine curve. More precisely, if the completion of the desingularization has 3 or more points at infinity, then Siegel's theorem says finitely many integer points. If there are one or two points at infinity, then one can us local/global criteria. Of course, one has to deal with the genus 0 curves that appear on a case-by-case basis. Sep 10 at 13:18
• Aren't rational points on the curves over number fields, resulting from the parametrization related to the N-conjecture?
– joro
Sep 10 at 14:39

Paper by L.D.Lehmer, "On a problem by Stormer", pp.57-79, Illinois J.Math. (received July 25, 1962; that's all the info about the paper that I have + a reprint) is related and should be certainly useful. It contains complete (extensive) tables of integers $$\ N\$$ such that the greatest prime divisor $$\ p|N\!\!\cdot\!(N-1)\$$ is $$\ p\le 41.\$$ This easily leads to a quick numerical method that would either discover all solutions with this kind of maximal prime $$\ p\$$ or would show that there are no new solutions but the two mentioned by OP (for such maximal primes $$\ \le 41$$).

A step in this direction can be as follows:

Let's consider the OP's equation with a minor shift of $$\ m\ n$$:

$$x^m\cdot(x+1)\ =\ y^n\cdot(y+1) \tag2$$ where $$x>y\qquad\text{and}\quad x\ m\ y\ n\,\in\mathbb N \tag3$$

The following function should be helpful, $$Q(t)\ :=\ \frac12\cdot\rho(t\cdot(t+1))$$ where $$\ \rho(s)\$$is defined as the product of all prime divisors of $$\ s;\$$ for instance $$\rho(1) = 1\quad\qquad \rho(8)=2\quad\qquad \rho(30)=\rho(900)=30$$ etc. Each solution of (2) implies a solution of equation

$$Q(x)\ =\ Q(y) \tag4$$

Thus an introductory and fundamental step toward solving (2) is

$$Q(t)\ =\ s \tag5$$

For the sake of (2) we need $$\ s\$$ that allows more than one solution $$\ t;\$$ there are relatively few of such cases, and when we have one then it is more often that not, that there is no respective solution of (2), a small bunch of elementary theorems can remove most of the negative cases.

For instance:

$$Q(2)=Q(3)=Q(8)\qquad=\qquad \mathbf3$$ leads to solution $$\ 3\cdot4\ =\ 2^2\cdot3\$$ of (2). Then we have

$$Q(5)=Q(9)=Q(15)=Q(24)=Q(80)\qquad=\qquad \mathbf{3\cdot5}$$ Let's look, say, at $$\ Q(5)=Q(24).\$$ If possible, it'd be induced by equation (2) of the following form:

$$24^m\cdot25\ =\ 5^n\cdot 6$$ This would force $$\ n=2\$$ hence $$\ 24^m\cdot25=5^2\cdot6,\$$ i.e. $$\ 24^m=6\$$ -- a contradiction hence no solution in this case. Another way to see it is that right away we would have m=1 -- a contradiction again.

Or, one more case for now, let's look at $$\ Q(9)=Q(80)\$$ hence at $$80^n\cdot 81\ =\ 9^m\cdot 10$$ Again, you can see that $$\ n=1\$$ and $$\ m=2\$$ -- a contradiction.

In order to gain extra efficiency, one needs some additional theorems.

• Many thanks for this excellent and very nice answer. I'm impressed, your answer and the answers of the other (MSE and MO) users could do, in my opinion, an interesting article in mathematics. Sep 12 at 10:01

EDIT This answer doesn't satisfy OP's assumptions. Currently keeping it because of existence of infinitely many solutions over extensions of the integers and the relation with N-conjecture.

If you set $$m=n$$ you have infinitely many solutions $$x=y$$.

If $$m=n=2$$, the solutions are $$x=y,x= -y-1$$.

If $$m=n=3$$ we have the factorization $$(-x + y) \cdot (x^2 + x*y + y^2 + x + y)$$. Maybe for deep reasons, the quadratic factor doesn't have solutions over the integers, but we believe it has infinitely many solutions over $$\mathbb{Z}[i]$$.

For $$m=n=4$$ we have a cubic factor, which is expected to be of genus $$1$$, and according to sage it is genus zero, which might give integral points.

For $$m=n < 30$$, the higher degree factor is genus zero, probably some algebraic geometer will explain it.

• OT's assumption was $\ x>y.\$ Thus $\ m<n\$ hence your discussion is a total miss. Furthermore, even when, against the OP's assumption, $\ m=n,\$ then the situation is totally trivial, $\ x=y.$ Somehow, your neglecting the assumption about all parameters being positive (even $\ \ge 2$). Thus, once again, this your second part is irrelevant to the problem. Sep 10 at 7:22
• We're talking here about positive real numbers (natural numbers). Just look at the inequalities. Sep 10 at 7:45
• @WlodAA Thanks, you are right.
– joro
Sep 10 at 7:46
• When I can I'm going to accept an answer for this post, after I read the contributions. Many thanks for your excellent answer, I think that this time I'm going to choose the other answer. Sep 10 at 12:01