Construction of a 8-degree polynomial with 16 real numbers (Vietnam TST 2016/6) Given $16$ distinct real numbers $\alpha_1,\alpha_2,\ldots,\alpha_{16}$. For each polynomial $P$, denote $$V(P)=P(\alpha_1)+P(\alpha_2)+\cdots+P(\alpha_{16}). $$Prove that there is a monic polynomial $Q$, $\deg Q=8$ satisfying:
i) $V(QP)=0$ for all polynomial $P$ has $\deg P<8$.
ii) $Q$ has $8$ real roots (including multiplicity).

I've managed to prove the existance of the coefficients of such a $Q$ but I do not know how to construct a $Q$ that only has real roots.
 A: I try a solution.
Let $E=\mathbb{R}_{8}[x]$, and define on $E$ 
$$<A,B>=\sum_{k=1}^{16}A(\alpha_k)B(\alpha_k)$$
We see easily that $<.,.>$ is a scalar product on $E$. 
Let $\{x^j, j=0,\cdots, 8\}$ the standard basis of $E$, and $A_j$, $j=0,\cdots,8$, the orthonormal basis deduced from this standard basis by Gram-Schmidt. 
Then the usual proof show that the $A_j$ have $j$ real roots (for $j\geq 1$). In fact this is trivial if $j=1$, and  if $A_j$, $j\geq 2$, has less than $j$ real roots, counting multiplicity, then it has a quadratic factor with no real roots, which has a constant sign, and we can find a non zero polynomial $B$, with degree $<j$, such that $B(x)A_j(x)\geq 0$ for all $x$. As $B$ is orthogonal to $A_j$, we get that $B(\alpha_k)A_j(\alpha_k)=0$ for $k=1,\cdots, 16$,and $BA_j=0$, a contradiction.
Now put $Q(x)=cA_8(x)$, with $c$ such that $Q$ is monic, and we are done.   
A: A Sketch of Proof
Let $W$ be the space of polynomials over $\mathbb{C}$ in variable $X$ of degree at most $8$.  Define an inner product $\langle\_,\_\rangle$ on $W$ via
$$\langle AB\rangle:=V\left(A\bar{B}\right)\,,$$
where $\bar{B}$ is the complex conjugate of $B$.  Prove that $\langle\_,\_\rangle$ is indeed an inner product.  Now, $W$ has a basis consisting of $T_i(X):=X^i$, for $i=0,1,\ldots,8$.  Perform a Gram-Schmidt ortonormalization on the $T_i$'s to obtain an orthonormal basis $\left\{S_0,S_1,\ldots,S_8\right\}$ with $S_i$ having degree $i$.  We may assume that $S_i(X)\in\mathbb{R}[X]$ for each $i$.  Then, take $Q$ to be the monic multiple of $S_8$.  Clearly, $Q$ satisfies the first condition.  
If $Q$ has less than $8$ real roots $r_1,\ldots,r_k$ with $k<8$ including multiplicities, then take $P$ as a follows:
$$P(X):=\prod_{j=1}^k\,\left(X-r_j\right)\,.$$
As $\langle Q,Q\rangle>0$, we conclude that there exists $i\in\{1,2,\ldots,16\}$ such that $\alpha_i\neq r_j$ for all $j=1,2,\ldots,k$.  By construction, $\langle Q,P\rangle=0$, but this is a contradiction as $Q(X)=P(X)\,R(X)$ for some $R(X)\in\mathbb{R}[X]$ with no real root (whence the values of $R\left(\alpha_i\right)$ are all positive or all negative).  In fact, it can be easily shown that all the roots of $Q$ are mutually distinct.
