, 2014

Suppose, for the sake of contradiction, that $f(x) = g(x)h(x)$ where $g(x)$ and $h(x)$ both have integer coefficients and degree at least one. Since
$$g(0)h(0) = f(0) = -3,$$
one of $g(0)$, $h(0)$ must be $\pm 1$, say $g(0) = \pm 1$. The absolute value of the product of the roots (in the complex numbers) of the polynomial $g(x)$ is $|g(0)| = 1$, so $g(z) = 0$ for some complex number $z$ with $|z| \le 1$. Now, $f(z) = 0$, so $z^n + z^{n-1} - z^{n-2} = 3$.
However, $|z^n + z^{n-1} - z^{n-2}| \le |z^n| + |z^{n-1}| + |z^{n-2}| \le 3$, with equality implying $|z| = 1$ and that $z^n, z^{n-1}$, and $-z^{n-2}$ all have the same argument. This is easily seen to be impossible, so we have reached the desired contradiction.