for $i=n$ we don't know if $|a_{n}|<1$ ?

something is missing on the problem, for $f=X^{n}$ $a)$ is not true.

oty wrote: for $i=n$ we don't know if $|a_{n}|<1$ ? Indeed, that is not given. In fact, it can be proved that it's quite the opposite. Also, I edited the problem. Now it is correct, with the non zero condition.

$f(X)=16(x-i/2)(x+i/2)(x-i/2)(x+i/2)(x-4)$ this polynomial is integer coefficients , there is four complexes roots not distincts i/2,i/2,-i/2,-i/2 with modulus strictly than 1 and one root 4 with modulus strictly greater than 1

Moubinool wrote: $f(X)=16(x-i/2)(x+i/2)(x-i/2)(x+i/2)(x-4)$ this polynomial is integer coefficients , there is four complexes roots not distincts i/2,i/2,-i/2,-i/2 with modulus strictly than 1 and one root 4 with modulus strictly greater than 1 $f$ must be monic. Catalin wrote: $f = \prod_{i=1}^n (X-a_i)$

Moubinool wrote: $f(X)=16(x-i/2)(x+i/2)(x-i/2)(x+i/2)(x-4)$ this polynomial is integer coefficients , there is four complexes roots not distincts i/2,i/2,-i/2,-i/2 with modulus strictly than 1 and one root 4 with modulus strictly greater than 1 At the first look I though it was stronger than X-2019 A part 2

a) Suppose $f$ is not irreducible, and let $f=gh$ for some $g,h\in\mathbb{Z}[X]$. Then either $g$ or $h$ will have all their roots less than $1$ in absolute value. Since the products of their roots are integers we obtain a contradiction. Hence $f$ is irreducible, and thus $a_1,...,a_n$ are indeed distinct. b) Firstly, let's prove the following claim. Claim: $|a_n|\geq 1$ Proof: If $|a_n|<1\Rightarrow |a_1...a_n|<1$, but from Viete $a_1...a_n\in\mathbb{Z^*}$, contradiction. $\square$ Now, suppose there are indices $i,j,k$, not all equal, for which $a_j^2=a_ia_k$. We will prove that $a_n^2=a_la_m$, for some $l<m$. Then, if $m\neq n$, $|a_n^2|=|a_la_m|<1$, false; if $m=n$ then $|a_n|=|a_l|<1$, which is again false. We can prove the above claim in two more or less equivalent ways. Method 1: Notice that $f$ being irreducible means that $f$ is the minimal polynomial of its roots. Claim: $a_i\neq -a_j$, $\forall i,j\in\{1,2,...,n\}$ Proof : Suppose $a_i=-a_j$ for some $i,j$. Hence $f(-a_i)=0\Rightarrow f(x)|f(-x)\Rightarrow f(-a_n)=0$, so $a_n=-a_k$ for some $k$. Since $a_n\neq 0$, $|a_n|=|-a_k|<1$, false. $\square$ From the second claim $j\neq k$. Let $g=\prod_{l<m}(X^2-a_la_m)$. Since $g$ is symetrical in $a_1,..., a_n$, $g\in\mathbb{Z}[X]$. As $g(a_j)=0\Rightarrow f|g\Rightarrow g(a_n)=0$, so $a_n^2=a_la_m$, for some $l<m$, as needed $\square$ Method 2: Let $G$ the Galois group of $f$. As this polynomial is irreducible, $G$ needs to be transitive. Then there exists $\sigma\in G$ with $\sigma(a_j)=a_n\Rightarrow a_n^2=\sigma(a_i)\sigma(a_k)=a_la_m$, for some $l<m$, as $\sigma$ permutes $f$'s roots. $\square$