Subfield: Number Theory


Question: Find all polynomials $P(x)$ of odd degree $d$ and with integer coefficients satisfying the following property: for each positive integer $n$, there exist $n$ positive integers $x_{1}, x_{2}, \ldots, x_{n}$ such that $\frac{1}{2}<\frac{P\left(x_{i}\right)}{P\left(x_{j}\right)}<2$ and $\frac{P\left(x_{i}\right)}{P\left(x_{j}\right)}$ is the $d$-th power of a rational number for every pair of indices $i$ and $j$ with $1 \leqslant i, j \leqslant n$.

Solution:
Let $P(x)=a_{d} x^{d}+a_{d-1} x^{d-1}+\cdots+a_{0}$. Consider the substitution $y=d a_{d} x+a_{d-1}$. By defining $Q(y)=P(x)$, we find that $Q$ is a polynomial with rational coefficients without the term $y^{d-1}$. Let $Q(y)=b_{d} y^{d}+b_{d-2} y^{d-2}+b_{d-3} y^{d-3}+\cdots+b_{0}$ and $B=\max _{0 \leqslant i \leqslant d}\left\{\left|b_{i}\right|\right\}$ (where $b_{d-1}=0$ ).

The condition shows that for each $n \geqslant 1$, there exist integers $y_{1}, y_{2}, \ldots, y_{n}$ such that $\frac{1}{2}<\frac{Q\left(y_{i}\right)}{Q\left(y_{j}\right)}<2$ and $\frac{Q\left(y_{i}\right)}{Q\left(y_{j}\right)}$ is the $d$-th power of a rational number for $1 \leqslant i, j \leqslant n$. Since $n$ can be arbitrarily large, we may assume all $x_{i}$ 's and hence $y_{i}$ 's are integers larger than some absolute constant in the following.

By Dirichlet's Theorem, since $d$ is odd, we can find a sufficiently large prime $p$ such that $p \equiv 2(\bmod d)$. In particular, we have $(p-1, d)=1$. For this fixed $p$, we choose $n$ to be sufficiently large. Then by the Pigeonhole Principle, there must be $d+1$ of $y_{1}, y_{2}, \ldots, y_{n}$ which are congruent $\bmod p$. Without loss of generality, assume $y_{i} \equiv y_{j}(\bmod p)$ for $1 \leqslant i, j \leqslant d+1$. We shall establish the following.

- Claim. $\frac{Q\left(y_{i}\right)}{Q\left(y_{1}\right)}=\frac{y_{i}^{d}}{y_{1}^{d}}$ for $2 \leqslant i \leqslant d+1$.

Proof. Let $\frac{Q\left(y_{i}\right)}{Q\left(y_{1}\right)}=\frac{l^{d}}{m^{d}}$ where $(l, m)=1$ and $l, m>0$. This can be rewritten in the expanded form

$$
b_{d}\left(m^{d} y_{i}^{d}-l^{d} y_{1}^{d}\right)=-\sum_{j=0}^{d-2} b_{j}\left(m^{d} y_{i}^{j}-l^{d} y_{1}^{j}\right)
\tag{1}
$$

Let $c$ be the common denominator of $Q$, so that $c Q(k)$ is an integer for any integer $k$. Note that $c$ depends only on $P$ and so we may assume $(p, c)=1$. Then $y_{1} \equiv y_{i}(\bmod p)$ implies $c Q\left(y_{1}\right) \equiv c Q\left(y_{i}\right)(\bmod p)$.

- Case 1. $p \mid c Q\left(y_{1}\right)$.

In this case, there is a cancellation of $p$ in the numerator and denominator of $\frac{c Q\left(y_{i}\right)}{c Q\left(y_{1}\right)}$, so that $m^{d} \leqslant p^{-1}\left|c Q\left(y_{1}\right)\right|$. Noting $\left|Q\left(y_{1}\right)\right|<2 B y_{1}^{d}$ as $y_{1}$ is large, we get

$$
m \leqslant p^{-\frac{1}{d}}(2 c B)^{\frac{1}{d}} y_{1} .
\tag{2}
$$

For large $y_{1}$ and $y_{i}$, the relation $\frac{1}{2}<\frac{Q\left(y_{i}\right)}{Q\left(y_{1}\right)}<2$ implies

$$
\frac{1}{3}<\frac{y_{i}^{d}}{y_{1}^{d}}<3
\tag{3}
$$

We also have

$$
\frac{1}{2}<\frac{l^{d}}{m^{d}}<2
\tag{4}
$$



Now, the left-hand side of (1) is

$$
b_{d}\left(m y_{i}-l y_{1}\right)\left(m^{d-1} y_{i}^{d-1}+m^{d-2} y_{i}^{d-2} l y_{1}+\cdots+l^{d-1} y_{1}^{d-1}\right) .
$$

Suppose on the contrary that $m y_{i}-l y_{1} \neq 0$. Then the absolute value of the above expression is at least $\left|b_{d}\right| m^{d-1} y_{i}^{d-1}$. On the other hand, the absolute value of the right-hand side of (1) is at most

$$
\begin{aligned}
\sum_{j=0}^{d-2} B\left(m^{d} y_{i}^{j}+l^{d} y_{1}^{j}\right) & \leqslant(d-1) B\left(m^{d} y_{i}^{d-2}+l^{d} y_{1}^{d-2}\right) \\
& \leqslant(d-1) B\left(7 m^{d} y_{i}^{d-2}\right) \\
& \leqslant 7(d-1) B\left(p^{-\frac{1}{d}}(2 c B)^{\frac{1}{d}} y_{1}\right) m^{d-1} y_{i}^{d-2} \\
& \leqslant 21(d-1) B p^{-\frac{1}{d}}(2 c B)^{\frac{1}{d}} m^{d-1} y_{i}^{d-1}
\end{aligned}
$$

by using successively (3), (4), (2) and again (3). This shows

$$
\left|b_{d}\right| m^{d-1} y_{i}^{d-1} \leqslant 21(d-1) B p^{-\frac{1}{d}}(2 c B)^{\frac{1}{d}} m^{d-1} y_{i}^{d-1},
$$

which is a contradiction for large $p$ as $b_{d}, B, c, d$ depend only on the polynomial $P$. Therefore, we have $m y_{i}-l y_{1}=0$ in this case.

- Case 2. $\left(p, c Q\left(y_{1}\right)\right)=1$.

From $c Q\left(y_{1}\right) \equiv c Q\left(y_{i}\right)(\bmod p)$, we have $l^{d} \equiv m^{d}(\bmod p)$. Since $(p-1, d)=1$, we use Fermat Little Theorem to conclude $l \equiv m(\bmod p)$. Then $p \mid m y_{i}-l y_{1}$. Suppose on the contrary that $m y_{i}-l y_{1} \neq 0$. Then the left-hand side of (1) has absolute value at least $\left|b_{d}\right| p m^{d-1} y_{i}^{d-1}$. Similar to Case 1, the right-hand side of (1) has absolute value at most

$$
21(d-1) B(2 c B)^{\frac{1}{d}} m^{d-1} y_{i}^{d-1},
$$

which must be smaller than $\left|b_{d}\right| p m^{d-1} y_{i}^{d-1}$ for large $p$. Again this yields a contradiction and hence $m y_{i}-l y_{1}=0$.

In both cases, we find that $\frac{Q\left(y_{i}\right)}{Q\left(y_{1}\right)}=\frac{l^{d}}{m^{d}}=\frac{y_{i}^{d}}{y_{1}^{d}}$.

From the Claim, the polynomial $Q\left(y_{1}\right) y^{d}-y_{1}^{d} Q(y)$ has roots $y=y_{1}, y_{2}, \ldots, y_{d+1}$. Since its degree is at most $d$, this must be the zero polynomial. Hence, $Q(y)=b_{d} y^{d}$. This implies $P(x)=a_{d}\left(x+\frac{a_{d-1}}{d a_{d}}\right)^{d}$. Let $\frac{a_{d-1}}{d a_{d}}=\frac{s}{r}$ with integers $r, s$ where $r \geqslant 1$ and $(r, s)=1$. Since $P$ has integer coefficients, we need $r^{d} \mid a_{d}$. Let $a_{d}=r^{d} a$. Then $P(x)=a(r x+s)^{d}$. It is obvious that such a polynomial satisfies the conditions.

Final answer: $P(x)=a(r x+s)^{d}$ where $a, r, s$ are integers with $a \neq 0, r \geqslant 1$ and $(r, s)=1$