Is it true that for every natural number $n \geq 3$, there exists an irreducible polynomial $f(x) = \sum_{i=0}^{n} a_{i}x^{i} \in\mathbb{Z}[x],$all of whose coefficients are non-zero and all of its roots are distinct irrational numbers?
Even for $n = 3$, I am finding it hard to prove. My attempt for $n = 3$ is as follows:
For $n = 3$, I figured that if $f(x) = ax^{3}+bx^{2}+cx+d$ then its discriminant is equal to $$D(f) := b^{2}c^{2}-4ac^{3}-4b^{3}d-27a^{2}d^{2}+18abcd.$$ The roots of $f(x)$ will all be real if $D(f) > 0$. So, the roots of $f(x)$ will be three distinct irrational when $D(f) > 0$ and $f(r) \neq 0$ for any $r$ in the following set $R$.
$$R:=\big\{ r = \frac{p}{q} : p \mid d \text{ and } q \mid a \big\}.$$
Since $(0, \infty)$ is open in $\mathbb{R}$, there must be an Euclidean open set worth of $(a,b,c,d)\in\mathbb{R}^{4}$ such that $D(f) > 0$. We call this Euclidean open set $O_{1}$. I do not see what else can be said about this $O_{1}$.
My strategy was to show that set $S$ of $(a,b,c,d)\in\mathbb{Q}^{4}$ such that $f(x)$ is irreducible is “quite large” in some sense such that it actually intersects with $O_{1}$. Then, we choose $(a,b,c,d)\in\mathbb{Q}^{4}\cap O_{1}$, which will correspond to a $f(x)$ which is irreducible and has three distinct irrational roots.
Comment- See even if my approach for $n=3$ works, this method would not work for higher degrees because $D(f)>0$ is not enough to have that all roots are irrational.
You can drop the condition that the coefficients are all nonzero: If $f(x) \in \mathbb{Z}[x]$ of degree $n \geq 2$ is irreducible and all of its roots are real (and hence irrational because $f$ is irreducible), then for any integer $N$, the polynomial $f(x + N)$ is also irreducible and has all roots real. For all but finitely many $N$, $f(x + N)$ will also have all coefficients nonzero.
Indeed, the $x^k$ coefficient of a polynomial is nonzero if and only if the $k$-th derivative of the polynomial has a nonzero value at zero, and the $k$-th derivative of $f(x + N)$ evaluated at zero is equal to the $k$-th derivative of $f(x)$ evaluated at $N$. So, we just need to choose $N$ so that $f(N), f'(N), f''(N), \ldots, f^{(n-1)}(N)$ are all nonzero, which means avoiding the roots of a finite list of polynomials, each of which has finitely many roots.
So, it suffices to find an irreducible integer polynomial of degree $n$ with all roots real. This is equivalent to finding a totally real number field of degree $n$. There are a number of ways of constructing totally real number fields of any given degree; this answer gives one such way.