Bump. Any solution?

See: http://math.stackexchange.com/questions/1872024/construction-of-a-8-degree-polynomial-with-16-real-numbers

does amyone have a bit elementary solution to this nice problem? The above is quite inaccessible for me.

Something is missing here. The exact problem is "Prove that there is only one polynomial...".

Who can solve it

We'll prove that there is a unique monic polynomial with degree 8 which satisfies the problem's conditions. Call polynomial $T\in \mathbb R[x]$ good if it is not a zero polynomial and for every polynomial $P$ with $\text{deg} \ P < 8$, $V(TP)=0$. $\textbf{Claim 1:}$ For every good polynomial $T$, $\text{deg} \ T\ge 8$. $\textit{Proof:}$ Suppose the contrary, there is a good polynomial $T$ with degree less than 8. Taking $P=T$, we have $V(T^2)=0$. $$T(\alpha_{1})^2+T(\alpha_{2})^2+\dots + T(\alpha_{16})^2=0$$ Which implies $T(\alpha_{1})=T(\alpha_{2})=\dots =T(\alpha_{16})=0$. As $\text{deg} \ P <8 <16$, then we must have that $T$ is a zero polynomial. Contradicting the definition of good polynomial. So, for every good polynomial $T$, $\text{deg} \ T\ge 8$. (Claim 1 is proven) $\textbf{Claim 2:}$ There is a monic good polynomial with degree 8. $\textit{Proof:}$ For every non-negative integer $m$, define $S_m=\sum_{i=1}^{16}\alpha_{i}^m$. For every $i=0,1,2,\dots,8$, define vector $v_i=(S_i,S_{i+1},\dots,S_{i+7})$. Because $v_0,v_1,v_2,\dots,v_8$ are all 8-dimensional vectors and there are 9 vectors, $v_0,v_1,v_2,\dots,v_8$ are linearly dependent. Therefore, there exist $u_0,u_1,\dots,u_8$ not all zero such that $$u_0v_0+u_1v_1+\dots + u_8v_8=0$$ If $u_8=0$, then $T(x)=u_7x^7+u_6x^6+\dots + u_0$ is a good polynomial. Contradicting claim 1. Therefore, $u_8\neq 0$. It is easy to check that $T(x)=x^8+\frac{u_7}{u_8}x^7+\frac{u_6}{u_8}x^6+\dots + \frac{u_0}{u_8}$ is a good polynomial. (Claim 2 is proven) $\textbf{Claim 3:}$ Monic good polynomial with degree 8 is unique. $\textit{Proof:}$ Suppose the contrary, there exists 2 different monic good polynomial with degree 8, $T_1(x)$ and $T_2(x)$. Notice that $R(x)=T_1(x)-T_2(x)$ must be a good polynomial too. However, as $T_1(x)$ and $T_2(x)$ are both monic, $\text{deg} \ R$ must be less than 8. Contradicting claim 1. (Claim 3 is proven) Now, define $Q$ as the only monic good polynomial with degree 8. $\textbf{Claim 4:}$ All roots of $Q$ are real. $\textit{Proof:}$ Assume the contrary, not all roots of $Q$ are real. Define $x_1,x_2,\dots,x_k$ as the root of $Q$ (counting multiplicities). Then, $k\le 7$. Define $L(x)$ such that $Q(x)=(x-x_1)(x-x_2)\dots (x-x_k)L(x)$. $L(x)$ doesn't have any real root, so either $L(x)>0 \ \forall \ x$ or $L(x)<0 \ \forall \ x$. Notice that if we take $P(x)=(x-x_1)(x-x_2)\dots (x-x_k)$, we have $V(QP)>0$ if $L(x)>0 \ \forall \ x$; and $V(QP)<0$ if $L(x)<0 \ \forall \ x$. Which is a contradiction as we must have $V(QP)=0$. So, all roots of $Q$ are real. (Claim 4 is proven) By claim 2,3, and 4, the problem is done