The question is in the title, $n>m>2$ are integers. All text below is the context.
Two weeks ago user759001 asked on integer solutions $x>y\ge 2$ of a Diophantine equation $$x^{m-1}(x+1)=y^{n-1}(y+1)\tag{1}$$ for integers $m,n\geq 2$. The only known solutions are $(x,y;m,n)=(3,2;2,3)$ and $(98,21;2,3)$. User2020201 showed that $m<n$. I conjectured that there are no solutions when $m|n$ and proved the conjecture in particual cases (when $(m,n)$ is $(2,6)$, $(3,19)$, or $(4,12)$. Also I guess I have a proof when $n=2m$), see this answer.
According to [G], Diophantine equations with two variables of degree greater than two have infinitely many (integer) solutions only in very rare cases. In particular, by a special and very complicated method K. Zigel’ (Siegel?) showed the following
Theorem. Let $P(x,y)$ be an irreducible polynomial of two variables with integer coefficients of a total degree greater than two (that is, $P(x,y)$ contains a monomial $ax^ky^s$, where $k+s>2$). (The irreducibility of $P(x,y)$ means that it cannot be represented as a product of two non-constant polynomials with integer coefficients). If an equation $P(x,y)=0$ has infinitely many integer solutions $(x,y)$ then there exist an integer $r$ and integers $a_i$, $b_i$ for each $-r\le i\le r$ such that if in the equation $P(x,y)=0$ we make a substitution $x=\sum_{i=-r}^r a_it^i$ and $y=\sum_{i=-r}^r b_it^i$ then we obtain an identity.
In order to apply this theorem to user759001’s equation for fixed $n>m>2$ we need irreducibility of the polynomial $y^n+y^{n-1}-x^m-x^{m-1}$. It looks plausible and easy to show, but, unfortunately, I am not a specialist in factorization of multivariable polynomials, so I decided to ask MSE community for help. Thanks.
References
[G] Gel’fand A.O. Solutions of equations in integer numbers, 3rd edn., Moscow, Nauka, 1978, in Russian.
from Hot Weekly Questions - Mathematics Stack Exchange
Alex Ravsky
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