An IMO shortlist polynomial problem:

Let $f$ and $g$ be two nonzero polynomials with integer coefficients and $\deg f>\deg g$. Suppose that for infinitely many primes $p$ the polynomial $pf+g$ has a rational root. Prove that $f$ has a rational root.

I could hardly make any real progress.What I could find was : I found using rational root theorem for large enough $p$ if the rational root of $pf+g$ is $r/s$ then $s \in \{1,p\}$

any kinds of helps are appreciated


Note: in this proof, I will extract a lot of times subsequences of the primes such that $pf+g$. When I then write something like “that sequence of $p$ is convergent”, it always refers to the extracted subsequence.

Let, for all such prime numbers, $\frac{r_p}{s_p}$ be a rational root (in irreducible form) of $pf+g$.

Thus for infinitely many $p$, $\frac{g}{f}\left(\frac{r_p}{s_p}\right)=-p \rightarrow -\infty$. But $g/f$ is a rational fraction of negative degree, so that $r_p/s_p$ is bounded.

By the rational root theorem, if $d$ is the dominant coefficient of $f$, and $f_0=f(0)$ (if $f_0=0$ we are done), $g_0=g(0)$, $s_p|pd$, $r_p|pf_0+g_0$, ie $pf_0+g_0=C_pr_p$.

  1. Assume that there are infinitely many $p$ such that $p|s_p$. Then there is a divisor $\delta$ of $d$ such that $s_p=p\delta$ infinitely many times.

Then for such $p$, $\frac{r_p}{s_p}=\frac{1}{C_p\delta}\frac{pf_0+g_0}{p}$.

1a. Assume that $C_p$ is unbounded, then there is a subsequence of $r_p/s_p$ going to zero, and thus it follows that $(g/f)(0)$ is undefined, so $f(0)=0$, which we assumed wasn’t the case.

1b. So there are infinitely many $p$ such that $s_p=p\delta$, and $C_p=N$ for some integer $N$.

Then for these $p$, $\frac{r_p}{s_p}=\frac{1}{N\delta}\frac{pf_0+g_0}{p}$. Thus $r_p/s_p \rightarrow \frac{f_0}{N\delta}=\alpha$. Thus $(g/f)(\alpha)$ is undefined and $f(\alpha)=0$.

  1. Otherwise, there is a $\delta$ with $s_p=\delta$ for infinitely many $p$.

Then $r_p=\delta\frac{r_p}{s_p}=\frac{pf_0+g_0}{C_p\delta}$ is bounded as well, so we can re-extract so that $r_p,s_p$ are constants (called $r,\delta$). This entails $f(r/\delta)=g(r/\delta)=0$ and we are done.

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