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7 votes
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Divisibility chains and polynomials

Every $P$ is a counterexample. Indeed, given a polynomial $P$ consider the recursive sequence $b_{n+1}=f(b_n)$ where I take $f(x)=x+P(x)$, say. Then $P(b_{n+1}) = P(b_n + P(b_n)) \equiv P(b_n) \equiv ...
1 vote

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

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 ...
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1 vote

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

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 ...
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8 votes
Accepted

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 ...

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