Finding an irreducible polynomial of degree $n$ in $\mathbb Q[X]$ with real roots Context.
I was trying to prove that for a given $n$, there exist a totally real number field of degree $n$.
I understood that it was equivalent to find a polynomial $P$ such that
(i) $P\in\mathbb Q[X]$ ;
(ii) $P$ is irreducible ;
(iii) $P$ only has real roots in $\mathbb C$ ;
(iv) $P$ has degree $n$.
I succeeded in my initial problem thanks to this and this two MSE questions.
But the construction is abstract, and I can not deduce from it an explicit polynomial satisfying the four conditions.
The question.

For a given $n$, do you know how can I construct an explicit polynomial $P$ satisfying (i), (ii), (iii) and (iv)?

 A: I am assuming that by "real roots in $\mathbb{C}$" you mean real roots (from $\mathbb{R}$). One possible family of polynomials that satisfy all these criteria (for $n\geq5$):
$$
f_n(x)=(x-1)(x-2)\cdots(x-n)+1.
$$
Proof.
Ad (i) and (iv) Trivial.
Ad (ii) This polynomial is well known to be irreducible, it follows nicely from irreducibility criterion of Pólya, see for example Prove that the polynomial $(x-1)(x-2)\cdots(x-n) + 1$, $ n\ge1 $, $ n\ne4 $ is irreducible over $\mathbb Z$ and/or linked questions.
Ad (iii) (sketch) This is a bit technical but can be done (holds for $n\geq 4$). Idea is that we have $f_n(i)=1$ for $i=1\dots n$. Now if we look at $f_n(k/2)$ for $k=1,3,\dots,2n-1$, we get another $n$ points, out of which $\lceil n/2 \rceil$ are negative. So by examining sign changes, we can reach the conclusion that $f(x)$ has exactly $n$ real roots.
Remark: The polynomial can be also slightly modified to avoid the limitation $n \geq 5$, for example $5\cdot (x-1)(x-2)\cdots(x-n)+1$ can be shown to satisfy all the conditions for $n \geq 1$.
A: Take an Eisenstein polynomial ( link ) $P(x)$ with integer coefficients for a prime $p$. Now consider $a$ integer such that $a+ Im(x_k)>0$ for all roots $x_k$ of $P(x)$ and moreover $a\equiv 0 \bmod{p^2}$. Then 


*

*$P(x- i a) \in \mathbb{Z}[x]$ has all the roots in the upper half plane. 

*$Re(P(x- i a))$ is  an Eisenstein polynomial for $p$.
Denote by $Q(x)=Re(P(x-ia))$.
From 1.  $Q(x)$ has all the roots real (and distinct) (see link )
From 2.  $Q(x)$ is irreducible over $\mathbb{Q}$. 
Example: $P(x) = x^n+3$, $\ \ p=3$, $a=9$.
$$
Q(x)=1/2(\ (x+9i)^n+(x-9i)^n)+3 $$ 
A: Take a polynomial with large integer coefficients with $n$ real roots.
Then perturb the coefficients slightly (add or subtract one from some
of them) to make the polynomial reduce modulo $2$ to an irreducible
polynomial modulo $2$. Your new polynomial is irreducible over $\Bbb Q$
and unless you have been unlucky or careless, will still have $n$ real
roots.
