Selection tests for the Gulf Mathematical Olympiad 2013

Assume that there exist two non-constant polynomials $g(X)$ and $h(X)$ with integer coefficients such that $f(X)=g(X)h(X)$. Because $p=g(0)h(0)$ is prime, we can assume that $|g(0)|=1$.

Because the modulus of the product of the complex roots of $g(X)$ is equal to $1$, at least one of these roots, say ${\omega }_0$, has modulus less than or equal to $1$. But $f\left({\omega }_0\right)=0$. We deduce that $$\begin{array}{rl}p& =\left|a_n{\omega }_0^n+a_{n-1}{\omega }_0^{n-1}+\cdots +a_1{\omega }_0\right|\\ & \le \left|a_n\right|\cdot {\left|{\omega }_0\right|}^n+\left|a_{n-1}\right|\cdot {\left|{\omega }_0\right|}^{n-1}+\cdots +\left|a_1\right|\cdot \left|{\omega }_0\right|\\ & \le \left|a_n\right|+\left|a_{n-1}\right|+\cdots +\left|a_1\right|\end{array}$$ which is a contradiction. Therefore, $f(X)$ is irreducible.