Problem
Find all polynomials $P(x)$ with integer coefficients such that:
(i) $P(0)=0$ , i.e. the constant term is $0$
(ii) $-1<P(x)<1$ for all $-1<x<1$
I made this problem while trying to solve the APMO 2018 Polynomials Problem.
APMO $2018$ Problem $5$
Find all polynomials $P(x)$ with integer coefficients such that for all real numbers $s$ and $t$ , if $P(s)$ and $P(t)$ are both integers, then $P(st)$ is also an integer.
Solution
WLOG, we take $P(x)$ to have positive leading coefficient. As $P(x)$ is a polynomial with integer coefficients, it is clear that for any integer $k>1
$, $P(k)$ is an integer. Moreover, if for any $x$, $P(x)$ is integer, then, $P(kx)$ is also an integer. We let $\deg(P(x)) = d$ to be $d \geqslant 1$ (trivial at constant polynomial). Now, we define:
$$Q(x) = P(kx) – k^d \cdot P(x)$$
We can see that $\deg(Q(x))<d$. Now, we claim that $\deg(Q(x))=0$. We find positive integer $M$ such that for any value $i>j>M$, $P(i)>P(j)$ , $Q(i) \geqslant Q(j)$ and $P(i)>Q(i)$. By the definition of $Q(x)$, we can see that for all integral outputs of $P(x)$, we have an integral output for $Q(x)$. If $\deg(Q(x)) \geqslant 1$, its growth rate must be atleast the same growth rate of $P(x)$. However, this is impossible as $\deg(Q(x))<\deg(P(x))$. Thus, $Q(x)=c$ for some real $c$.
From this, it is clear that $P(x) = ax^d+b$ for integers $a,b$. It is clear that we can take $b$ to be any integer and WLOG, $a$ to be positive, thus we fix $b=0$ to find possible values for $a$. As $P(x) = ax^d$, we can note that $a^{-\frac{1}{d}}$ will produce an integral output. Then, we can take $s=t=a^{-\frac{1}{d}}$ and conclude that $P(a^{-\frac{2}{d}})$ is an integer. Thus, $\frac{1}{a}$ is an integer which would show that $a = \pm 1$. Thus, the answer is:
$$P(x) = \pm x^d + c$$
for non-negative integer $d$ and any integer $c$.
Thought for Alternate approach
WLOG, we can fix the constant term of $P(x)$ as 0. Assume that there is a value $-1<x<1, x \neq 0$ such that $P(x)$ is integer. Then, we can see that $P(x^k)$ is an integer for any positive integer $k$. Thus, we get infinitely many integral outputs in a finite area. As there is a local maxima and local minima, there are only finitely many possible integral outputs. Thus, one of these outputs must be an output infinitely often. Let that integer be $M$. Then, we can see that the polynomial $Q(x)=P(x)-M$ has infinitely many roots. Contradiction.
This tells us that $P(x)$ has no integral values between $-1<x<1$ except $0$. As we have fixed the constant term as $0$, we know that $P(0)=0$. Thus, we can see that for all values $-1<x<1$, we have $-1<P(x)<1$, which brings us to our question of finding all such polynomials. One polynomial other than $P(x) = \pm x^d + c$ which satisfy these conditions is for example, $P(x)=x^3+x^2-x$.
We could ask ourselves a weaker version of the same problem, specific to the APMO problem. We know that there are no integral values for $P(x)$ in $(-1,1)$ except at $x=0$. Thus, a weaker version of the problem would be to solve the same, with the additional condition that $P(x)$ has no roots in $(-1,1)$ except at $x=0$. Here, polynomials such as $P(x)=x^3+x^2-x$ fail to satisfy the conditions as one root is $x=\phi-1$ where $\phi$ is the golden ratio.
Thus, an alternative question would be to find all polynomials $P(x)$ with integer coefficients such that:
(i) For $x \in (-1,1)$ ; $P(x)=0 \implies x=0$
(ii) $-1<P(x)<1$ for all $-1<x<1$
Any help / suggestions / ideas are welcome.
Best Answer
I like your approach, but, unfortunately, there is a lot such polynomials. Denote their set by $\mathcal P$.
For a polynomial $p(x)$ let $\|p(x)\|=\sup\{|p(x)|:-1\le x\le 1\} $.
Let $p_0(x)=(x-1)x(x+1)$. Then $\alpha=\|p_0(x)\|=\frac {2\sqrt{3}}{9}\simeq 0.385.$.
Let $p$ be any polynomial with integer coefficients. Since $p$ is continuous and the set $[-1,1]$ is compact, $M=\|p(x)\|$ is finite. Pick $n$ such that $\alpha^nM<1$. Then $\|(p_0(x))^kp(x)\|<1$, so $(p_0(x))^kp(x)\in\mathcal P$. In particular, $xT_n(x)\in\mathcal P$, where $T_n(x)$ are Chebyshev_polynomials of the first kind. They have integer coefficients and $\|T(x)\|=1$. It seems that a polynomial $U_7(x)/8=16x^7-24x^5+10x^3-x$ also belongs to $\mathcal P$.
At last, remark that a product of any two polynomials from $\mathcal P$ again belongs to $\mathcal P$.