Problem 1012

Proof: Let $f(x) = g(x) h(x)$, where $g, h$ are polynomials with integer coefficients, not constants, and the leading coefficients are 1, then $x^{n} \equiv g(x) h(x) (\bmod 2)$. Therefore, we can set
$$\begin{array}{l} g(x) = x^{r} + 2\left(b_{r-1} x^{r-1} + \cdots + b_{0}\right), h(x) = \\ x^{n-r} + 2\left(c_{n-r-1} x^{n-r-1} + \cdots + c_{0}\right), 1 \leqslant r < n . \end{array}$$

By comparing the constant terms, we get
$$\begin{aligned} b_{0} & = c_{0} = \pm 1, x^{r} \cdot 2\left(c_{n-r-1} x^{n-r-1} + \cdots + c_{0}\right) \\ + x^{n-r} & \cdot 2\left(b_{r-1} x^{r-1} + \cdots + b_{0}\right) \equiv 0 (\bmod 4) . \end{aligned}$$

By comparing the coefficients of the lowest degree terms, we know $r = n-r$, i.e., $n = 2r$. Thus,
$$\left(c_{r-1} + b_{r-1}\right) x^{r-1} + \cdots + \left(b_{1} + c_{1}\right) x \equiv$$
$0 (\bmod 2), b_{1} + c_{1} \equiv 0 (\bmod 2)$. The coefficient of the linear term in $g(x) h(x)$ is $4\left(b_{1} c_{0} + c_{1} b_{0}\right) = 4 c_{0}\left(b_{1} + c_{1}\right) \equiv$ $0 (\bmod 8)$, which contradicts the fact that the coefficient of the linear term in $f$ is 4.