# If f\in \mathbb{Z}[X]f\in \mathbb{Z}[X] has the property that |f(x)|<1, \forall x\in (-2, 2)|f(x)|<1, \forall x\in (-2, 2), then prove that f=0f=0.

Let $$f\in \mathbb{Z}[X]f\in \mathbb{Z}[X]$$ such that $$|f(x)|<1, \forall x\in (-2, 2)|f(x)|<1, \forall x\in (-2, 2)$$. Prove that $$f=0f=0$$.
I couldn't make too much progress on this problem. I tried considering $$f=a_nX^n+a_{n-1}X^{n-1}+...+a_1X+a_0f=a_nX^n+a_{n-1}X^{n-1}+...+a_1X+a_0$$ and by setting $$x=0x=0$$ in the hypothesis I got that $$|a_0|<1|a_0|<1$$ and this doesn't look useful.
Then I thought about looking at $$ff$$'s degree, but I couldn't make any observations on this.
I believe that the key of the problem should be that the polynomial's coefficients are integers, but I don't know how to use that. Apart from the Rational Root Theorem (which doesn't seem useful here) I don't have in mind other results regarding polynomials with integer coefficients.

I can provide you with an answer based on Chebyshev polynomials. Suppose that $$ff$$ satisfies the stated assumptions, let $$kk$$ be the degree of $$ff$$, such that we have $$a_k \neq 0a_k \neq 0$$ and define $$g(x) := f(x)/a_kg(x) := f(x)/a_k$$. As $$a_k \in \mathbb{Z}a_k \in \mathbb{Z}$$ we must have $$|a_k|\geq 1|a_k|\geq 1$$ and therefore $$|g(x)| = |f(x)/a_k| < 1|g(x)| = |f(x)/a_k| < 1$$ for $$x \in (-2,2)x \in (-2,2)$$. Define the $$nn$$-th Chebyshev polynomial $$T_n(x) := \cos(n\arccos(x)) T_n(x) := \cos(n\arccos(x))$$ on $$[-1,1][-1,1]$$. Note that $$2^{1-n}T_n(x)2^{1-n}T_n(x)$$ has minimal supremum norm among all monic polynomials of degree $$nn$$ on $$[-1,1][-1,1]$$ for $$n \geq 1n \geq 1$$ (see here). By a rescaling argument, it follows that $$2T_n(x/2)2T_n(x/2)$$ has minimal supremum norm among all monic polynomials of degree $$nn$$ on $$[-2,2][-2,2]$$ for $$n \geq 1n \geq 1$$. It is easy to see from the definition that for $$n \geq 1n \geq 1$$ it holds that
$$\sup_{x \in [-2,2]} 2|T_n(x/2)| = 2.\sup_{x \in [-2,2]} 2|T_n(x/2)| = 2.$$
By continuity of $$gg$$, we have $$\sup_{x \in [-2,2]} |g(x)| \leq 1,\sup_{x \in [-2,2]} |g(x)| \leq 1,$$
and therefore we must have $$k < 1k < 1$$, as otherwise we would have found a monic polynomial with smaller supremum norm than the Chebyshev polynomial of the corresponding degree, a contradiction. It follows that $$gg$$ is constant and therefore equal to $$\frac{a_0}{a_k}\frac{a_0}{a_k}$$. Because of $$|f(0)| < 1|f(0)| < 1$$ we must have $$|a_0| < 1|a_0| < 1$$ and as $$a_0 \in \mathbb{Z}a_0 \in \mathbb{Z}$$ it follows that $$a_0 = 0a_0 = 0$$, therefore $$g \equiv 0 g \equiv 0$$ and hence $$f \equiv 0f \equiv 0$$, which contradicts the assumption that $$a_k \neq 0a_k \neq 0$$. Therefore, the only polynomial satisfying the hypotheses is the zero polynomial.