# If $f$ and $g$ are nonzero polynomials with $\deg f>\deg g$, and if $pf+g$ has a rational root for infinitely-many primes $p$, then …

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.