Subfield: Algebra


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

Solution:
Since $\operatorname{deg} f>\operatorname{deg} g$, we have $|g(x) / f(x)|<1$ for sufficiently large $x$; more precisely, there is a real number $R$ such that $|g(x) / f(x)|<1$ for all $x$ with $|x|>R$. Then for all such $x$ and all primes $p$ we have

$$
|p f(x)+g(x)| \geq|f(x)|\left(p-\frac{|g(x)|}{|f(x)|}\right)>0
$$

Hence all real roots of the polynomials $p f+g$ lie in the interval $[-R, R]$.

Let $f(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\cdots+a_{0}$ and $g(x)=b_{m} x^{m}+b_{m-1} x^{m-1}+\cdots+b_{0}$ where $n>m, a_{n} \neq 0$ and $b_{m} \neq 0$. Upon replacing $f(x)$ and $g(x)$ by $a_{n}^{n-1} f\left(x / a_{n}\right)$ and $a_{n}^{n-1} g\left(x / a_{n}\right)$ respectively, we reduce the problem to the case $a_{n}=1$. In other words one can assume that $f$ is monic. Then the leading coefficient of $p f+g$ is $p$, and if $r=u / v$ is a rational root of $p f+g$ with $(u, v)=1$ and $v>0$, then either $v=1$ or $v=p$.

First consider the case when $v=1$ infinitely many times. If $v=1$ then $|u| \leq R$, so there are only finitely many possibilities for the integer $u$. Therefore there exist distinct primes $p$ and $q$ for which we have the same value of $u$. Then the polynomials $p f+g$ and $q f+g$ share this root, implying $f(u)=g(u)=0$. So in this case $f$ and $g$ have an integer root in common.

Now suppose that $v=p$ infinitely many times. By comparing the exponent of $p$ in the denominators of $p f(u / p)$ and $g(u / p)$ we get $m=n-1$ and $p f(u / p)+g(u / p)=0$ reduces to an equation of the form

$$
\left(u^{n}+a_{n-1} p u^{n-1}+\ldots+a_{0} p^{n}\right)+\left(b_{n-1} u^{n-1}+b_{n-2} p u^{n-2}+\ldots+b_{0} p^{n-1}\right)=0 .
$$

The equation above implies that $u^{n}+b_{n-1} u^{n-1}$ is divisible by $p$ and hence, since $(u, p)=1$, we have $u+b_{n-1}=p k$ with some integer $k$. On the other hand all roots of $p f+g$ lie in the interval $[-R, R]$, so that

$$
\begin{gathered}
\frac{\left|p k-b_{n-1}\right|}{p}=\frac{|u|}{p}<R, \\
|k|<R+\frac{\left|b_{n-1}\right|}{p}<R+\left|b_{n-1}\right| .
\end{gathered}
$$

Therefore the integer $k$ can attain only finitely many values. Hence there exists an integer $k$ such that the number $\frac{p k-b_{n-1}}{p}=k-\frac{b_{n-1}}{p}$ is a root of $p f+g$ for infinitely many primes $p$. For these primes we have

$$
f\left(k-b_{n-1} \frac{1}{p}\right)+\frac{1}{p} g\left(k-b_{n-1} \frac{1}{p}\right)=0
$$

So the equation

$$
f\left(k-b_{n-1} x\right)+x g\left(k-b_{n-1} x\right)=0
\tag{1}
$$

has infinitely many solutions of the form $x=1 / p$. Since the left-hand side is a polynomial, this implies that (1) is a polynomial identity, so it holds for all real $x$. In particular, by substituting $x=0$ in (1) we get $f(k)=0$. Thus the integer $k$ is a root of $f$.

In summary the monic polynomial $f$ obtained after the initial reduction always has an integer root. Therefore the original polynomial $f$ has a rational root.
Analogously to the first solution, there exists a real number $R$ such that the complex roots of all polynomials of the form $p f+g$ lie in the disk $|z| \leq R$.

For each prime $p$ such that $p f+g$ has a rational root, by GAUSS' lemma $p f+g$ is the product of two integer polynomials, one with degree 1 and the other with degree $\operatorname{deg} f-1$. Since $p$ is a prime, the leading coefficient of one of these factors divides the leading coefficient of $f$. Denote that factor by $h_{p}$.

By narrowing the set of the primes used we can assume that all polynomials $h_{p}$ have the same degree and the same leading coefficient. Their complex roots lie in the disk $|z| \leq R$, hence VIETA's formulae imply that all coefficients of all polynomials $h_{p}$ form a bounded set. Since these coefficients are integers, there are only finitely many possible polynomials $h_{p}$. Hence there is a polynomial $h$ such that $h_{p}=h$ for infinitely many primes $p$.

Finally, if $p$ and $q$ are distinct primes with $h_{p}=h_{q}=h$ then $h$ divides $(p-q) f$. Since $\operatorname{deg} h=1$ or $\operatorname{deg} h=\operatorname{deg} f-1$, in both cases $f$ has a rational root.
Like in the first solution, there is a real number $R$ such that the real roots of all polynomials of the form $p f+g$ lie in the interval $[-R, R]$.

Let $p_{1}<p_{2}<\cdots$ be an infinite sequence of primes so that for every index $k$ the polynomial $p_{k} f+g$ has a rational root $r_{k}$. The sequence $r_{1}, r_{2}, \ldots$ is bounded, so it has a convergent subsequence $r_{k_{1}}, r_{k_{2}}, \ldots$ Now replace the sequences $\left(p_{1}, p_{2}, \ldots\right)$ and $\left(r_{1}, r_{2}, \ldots\right)$ by $\left(p_{k_{1}}, p_{k_{2}}, \ldots\right)$ and $\left(r_{k_{1}}, r_{k_{2}}, \ldots\right)$; after this we can assume that the sequence $r_{1}, r_{2}, \ldots$ is convergent. Let $\alpha=\lim _{k \rightarrow \infty} r_{k}$. We show that $\alpha$ is a rational root of $f$.

Over the interval $[-R, R]$, the polynomial $g$ is bounded, $|g(x)| \leq M$ with some fixed $M$. Therefore

$$
\left|f\left(r_{k}\right)\right|=\left|f\left(r_{k}\right)-\frac{p_{k} f\left(r_{k}\right)+g\left(r_{k}\right)}{p_{k}}\right|=\frac{\left|g\left(r_{k}\right)\right|}{p_{k}} \leq \frac{M}{p_{k}} \rightarrow 0
$$

and

$$
f(\alpha)=f\left(\lim _{k \rightarrow \infty} r_{k}\right)=\lim _{k \rightarrow \infty} f\left(r_{k}\right)=0
$$

So $\alpha$ is a root of $f$ indeed.

Now let $u_{k}, v_{k}$ be relative prime integers for which $r_{k}=\frac{u_{k}}{v_{k}}$. Let $a$ be the leading coefficient of $f$, let $b=f(0)$ and $c=g(0)$ be the constant terms of $f$ and $g$, respectively. The leading coefficient of the polynomial $p_{k} f+g$ is $p_{k} a$, its constant term is $p_{k} b+c$. So $v_{k}$ divides $p_{k} a$ and $u_{k}$ divides $p_{k} b+c$. Let $p_{k} b+c=u_{k} e_{k}$ (if $p_{k} b+c=u_{k}=0$ then let $e_{k}=1$ ).

We prove that $\alpha$ is rational by using the following fact. Let $\left(p_{n}\right)$ and $\left(q_{n}\right)$ be sequences of integers such that the sequence $\left(p_{n} / q_{n}\right)$ converges. If $\left(p_{n}\right)$ or $\left(q_{n}\right)$ is bounded then $\lim \left(p_{n} / q_{n}\right)$ is rational.

Case 1: There is an infinite subsequence $\left(k_{n}\right)$ of indices such that $v_{k_{n}}$ divides $a$. Then $\left(v_{k_{n}}\right)$ is bounded, so $\alpha=\lim _{n \rightarrow \infty}\left(u_{k_{n}} / v_{k_{n}}\right)$ is rational.

Case 2: There is an infinite subsequence $\left(k_{n}\right)$ of indices such that $v_{k_{n}}$ does not divide $a$. For such indices we have $v_{k_{n}}=p_{k_{n}} d_{k_{n}}$ where $d_{k_{n}}$ is a divisor of $a$. Then

$$
\alpha=\lim _{n \rightarrow \infty} \frac{u_{k_{n}}}{v_{k_{n}}}=\lim _{n \rightarrow \infty} \frac{p_{k_{n}} b+c}{p_{k_{n}} d_{k_{n}} e_{k_{n}}}=\lim _{n \rightarrow \infty} \frac{b}{d_{k_{n}} e_{k_{n}}}+\lim _{n \rightarrow \infty} \frac{c}{p_{k_{n}} d_{k_{n}} e_{k_{n}}}=\lim _{n \rightarrow \infty} \frac{b}{d_{k_{n}} e_{k_{n}}}
$$

Because the numerator $b$ in the last limit is bounded, $\alpha$ is rational.