Let $P$ and $Q$ be monic polynomials with integer coefficients and degrees $n$ and $d$ respectively, where $d\mid n$. Suppose there are infinitely many pairs of positive integers $(a,b)$ for which $P(a)=Q(b)$.

I would like to determine if exists a polynomial $R$ with integer coefficients such that $$P(x)=Q(R(x))$$

The second half of polynomials such that $P(k)=Q(l)$ for all integer $k$ is related though the condition here is weaker. I suspect the answer is yes (for polynomials, I've seen often that if some property occurs infinitely often then it occurs always).

My guess would be that we somehow construct a polynomial related to $P$ and $Q$ that ends up having infinitely many roots because of the infinitely many pairs $(a,b)$, so that we can force $P$ to conform to some sort of polynomial in $Q$. I'm not quite sure what to make of the $d\mid n$ condition; perhaps this could be strengthened? I haven't been able to find a counterexample that forces the divisibility.

  • $\begingroup$ Not sure if helpful at all but some ideas here. (1) Let the coefficient be $p_1,p_2,...,p_n$ and $q_1,q_2,...q_d$ respectively, then if we treat these coefficients as variables we essentially have $n+d$ unknowns but infinitely many equations. This could make use of the infinite pairs condition. (2) For the construction thing, it could be helpful to define $Q^{-1}$ in some way, and indeed that is true in some small cases where $Q^{-1}$ is calculable, but I wasn't able to proceed any further than that. $\endgroup$
    – cr001
    Aug 9, 2020 at 16:57
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    $\begingroup$ Lastly, the divisibility condition is necessary. $P(x)=x^3$ and $Q(x)=x^2$ will give a counter example: no $R$ exists and infinitely many integer pairs $P(n^2)=Q(n^3)$ exist. $\endgroup$
    – cr001
    Aug 9, 2020 at 17:00
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    $\begingroup$ I think that by Faltings' theorem, the curve $P(x)=Q(y)$ must have genus $\le1$. By Siegel's theorem the same holds in genus $1$. That leaves genus zero. Don't know for sure, but I suspect this to be true, the function field is rational after all, and that should/could help. Hmm. I should look at the curve one component at a time. And let's keep in mind Carl's (now deleted) answer as well. $\endgroup$ Aug 14, 2020 at 12:42
  • $\begingroup$ I checked out old suggestions to the Pearl Dive. This question was endorsed by Will Jagy. Unfortunately I forgot to act on it in a timely manner. $\endgroup$ Apr 24, 2022 at 6:19
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    $\begingroup$ Anyway, interesting related contributions are all eligible for the bounty. Sil's answer may be difficult to beat, but let's have a jam session. $\endgroup$ Apr 24, 2022 at 6:20

1 Answer 1


Todd Cochrane's article The Diophantine Equation $f(x) = g(y)$ provides a theorem which guarantees existence of a rational polynomial $R(x)$. Specifically if \begin{align} P(x) \equiv a_nx^n&+a_{n-1}x^{n-1}+\dots+a_0\\ &=b_my^m+b_{m-1}y^{m-1}+\dots+b_0 \equiv Q(y),\tag{*} \end{align} then the following is true (proof can be found in the article):

Suppose that $m \mid n$ and that $(a_n/b_m)$ is the $m$-th power of a rational number. Then either

  1. $P(x)=Q(R(x))$ for some polynomial $R(x)$ with rational coefficients taking integral values at infinitely many integers; or
  2. equation $(*)$ has at most finitely many integral solutions.

In our case $a_n/b_m=1$ is $m$-th power of rational number $1$. Also by assumption, we have infinitely many solutions of equation $(*)$, so the above gives us $P(x)=Q(R(x))$ with $R(x)$ over rationals. Furthermore since $Q(x)$ is monic, $R(x)$ cannot have a non-integral coefficient - that would either force some of the coefficients of $P(x)$ to be non-integral as well (see this nice proof from Doctor Who), or $Q(x)$ to be a constant polynomial (in which case we can trivially choose any $R(x)$ anyway).

So in any case, under given conditions existence of desired polynomial $R(x)$ over integers follows.


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