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## Irreductible polynomial 1

Prove that if $a \in \mathbb{Z}^*$ and $n\geq 2$ then the polynomial $f(x)=x^n+ax^{n-1}+...+ax-1$ is irreductible in $\mathbb{Z}[X]$.

Solution: Using the following Lemma we see that $f$ is irreductible if and only $-x^nf(\frac{1}{x})=x^n-ax^{n-1}-...-ax-1$ is irreductible. Therefore, we can consider $a\geq 1$. Since $f(0)f(1)=1-(n-1)a\leq 0$ we see that $f$ has a root $\alpha \in (0,1)$.
Let $f(x)=(x-\alpha)f_1(x)$, and inductively we get $f_1(x)=x^{n-1}+(a+\alpha)x^{n-2}+(a+\alpha a +\alpha^2)x^{n-3}+...+\frac{1}{a}$. It is easy to see that the coefficients of $f_1$ satisfy the hypothesis of This Problem so the roots of $f_1$ have modulus strictly greater than 1, or $f_1$ has a root $z$ which is a root of unity of some order $k$. Then $f$ has $z$ as a root, and since $f$ has integer coefficients, $1+x+...+x^{k-1}| x^{n}+...+x+1+(a-1)(x^{n-1} +...+x+1) - (a+1)$. If the remainder $r$ of $n+1$ modulo $k$ is not zero, then we reach a contradiction. Therefore $1+x+...+x^{k-1}| (a-1)(x^{n-1} +...+x+1) - (a+1)$ which implies $1+x+...+x^{k-1}|(a-1)x^{n}+a+1$. Therefore $z$ is a root for $(a-1)x^{n}+a+1$ ( then $a\neq 1$ ). Then $1=|z|^n=\frac{a+1}{a-1}$. Contradiction. Therefore $f_1$ has only roots of modulus strictly greater than 1.
Suppose now that $f$ is not irreductible. Then $f=gh,\ h,g \in \mathbb{Z}[X],\ \deg(g),\deg(h) \geq 1$. Now, at least one of $g,h$ has all its roots with modulus greater than 1. Suppose without loss of generality that $g$ has this property. Then $|g(0)|=\displaystyle \prod_{g(z)=0} |z| >1$. But, on the other hand $|h(0)g(0)|=1$ so $|g(0)|=1$. The contradiction obtained, proves that in fact, $f$ is irreductible.