Rational roots of polynomials Can one construct a sequence $(a_k)_{k\geqslant 0}$ of rational numbers such that, for every positive integer $n$ the polynomial $a_nX^n+a_{n-1}X^{n-1}+\cdots +a_0$ has exactly $n$ distinct rational roots ?
If we cannot construct it explicitly, can we show that such a sequence exists? 
PS: One can show (not easily) that the polynomial $\displaystyle \sum_{k=0}^n\frac{1}{3^{k^2}}X^k$ has $n$ distinct real roots.
 A: This does not answer the OP's question, rather gives a partial result. Anyhow, it is too long to fit into a comment. From now on, I work with the Zariski topology.
Let $p(x)=a_0+a_1x+\cdots+a_nx^n=(x-x_1)\cdots(x-x_n)\in\mathbb{Q}[x]$.
First, let us describe the set of coefficients $(a_0,\ldots,a_n)\in\mathbb{A}^{n+1}_{\mathbb{Q}}$ for which $p$ has distinct rational roots. Relations between the roots and coefficients (Vieta's formulae) tell us that 
$$a_i=(-1)^{(n-i)}a_ns_{n-i}(x_1,\ldots,x_n)$$
for $i=0,\ldots,n-1$, where $s_j$ denotes the $j$-th elementary symmetric polynomial. Now, let $X=Y=\mathbb{A}^{n+1}_{\mathbb{Q}}$, and consider the regular map $\varphi:X\longrightarrow Y$ defined by
$$\varphi(x_1,\ldots,x_n,a_n)=(a_0,\ldots,a_n).$$
I claim that the set of coefficients we are looking for is precisely $\varphi(X)\cap D(\Delta_p)\cap D(a_n)$, where $\Delta_p$ denotes the discriminant of $p$. Indeed, while $\varphi(X)$ guarantees rational roots, $D(\Delta_p)$ that they are distinct, $D(a_n)$ ensures that the leading coefficient is not zero.
Lemma. The map $\varphi:X\longrightarrow Y$ is dominant, that is, $\overline{\varphi(X)}=Y$.
Proof. We argue by contradiction. Suppose that $Z\subset Y$ is a proper closed subset containing $\varphi(X)$. Without loss of generality, we may assume that $Z=V(f)$ for some non-zero polynomial $f\in\mathbb{Q}[y_0,\ldots,y_n]$. Define $f_{a_n}=f(y_0,\ldots,y_{n-1},a_n)$. Now, by assumption, $f\circ\varphi$ vanishes identically on $X$, but then $f_{a_n}=0$ for all $0\neq a_n\in\mathbb{Q}$ since (and this is the key point) the elementary symmetric polynomials are algebraically independent over $\mathbb{Q}$. This implies that $f=0$, a contradiction.
As a dominant map, $\varphi$ contains a non-empty open set $U\subset Y$ in its image (the proof uses Noether normalisation). Therefore, $U\cap D(\Delta_p)\cap D(a_n)$ is a non-empty open set of coefficients for which $p$ has distinct rational roots. Let us denote this set by $D_n$, emphasising the degree of the polynomial. Now, if $(a_0,\ldots,a_n)$ is in the non-empty open set
$$V_n=\bigcap_{k=0}^n D_k\times\mathbb{A}^{n-k}_{\mathbb{Q}}\subset\mathbb{A}^{n+1}_{\mathbb{Q}},$$
then $p_k(x)=a_0+a_1x+\cdots+a_kx^k$ has distinct rational roots for all $k=0,\ldots,n$.
Therefore, in degree $n$, we have very many solutions.
Unfortunately, it is not clear how we can construct an infinite sequence in this way since $V_n$ might have empty intersection with the closed set $\{a_0\}\times\cdots\times\{a_{n-1}\}\times\mathbb{A}^1_{\mathbb{Q}}$ for a given $(a_0,\ldots,a_{n-1})\in V_{n-1}$. If we could bound the dimension of the complement of an open subset of $\varphi(X)\cap D(\Delta_p)\cap D(a_n)$ independently of $n$, then one could argue inductively. However, all we know for sure is that $\dim(U^c)\leq n-1$.
EDIT. I have just realised that, though it is dominant, $\varphi$ might not have any non-empty open set in its image, for $\mathbb{Q}$ is not algebraically closed. Unfortunately, this means that the conclusion I drew is wrong. However, the first part still makes sense.
