Find $P(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0 ,\ n\ge 1$ has $n$ roots $x_1,x_2,\ldots,x_n \le -1$ and such that $a_0^2+a_1a_n=a_n^2+a_0a_{n-1}.$ 
Let $P(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0 ,\  n\ge 1$ have $n$ roots $x_1,x_2,\ldots,x_n \le -1$ and  $a_0^2+a_1a_n=a_n^2+a_0a_{n-1}$.
  Find all such $P(x)$. (Poland 1990).

I used Viete Theorem and get: $$(x_1x_2\cdots x_n)^2 \pm \sum_{j=1}^n\prod_{i \not=j} x_i=1\pm(x_1x_2 \cdots x_n)(x_1+x_2+\cdots+x_n)$$ but not succeeded.
 A: Unless I misunderstand the problem,
$P(x) = x^2+2x+1$ meets the condition: Both roots are $-1$ and
$$a_0^2+a_1a_2 = a_2^2 + a_0a_1=3$$
Was the problem to find all such $P(x)$?
A: Let $s_i$ denote the elementary symmetric polynomials in $x_1,x_2,...,x_n$.
Then $a_0/a_n=(-1)^ns_n$, $a_1/a_n=(-1)^{n-1}s_{n-1}=(-1)^{n-1}s_n\left(\sum_{k=1}^{n}\frac{1}{x_k}\right)$, and $a_{n-1}/a_n=-s_1$.
We can re-interpret $a_0^2+a_1a_n=a_n^2+a_0a_{n-1}$ as
$$s_n^2+(-1)^{n-1}s_{n-1}=1-s_ns_{n-1}$$
$$s_n(s_n-s_{n-1})+(-1)^{n-1}s_{n-1}=1$$
Assume $n$ is even. Then, $s_n$ is positive, $-s_{n-1}$ is positive, $(-1)^{n-1}s_{n-1}$ is positive. If at least one of the $x_k<-1$ then $(-1)^{n-1}s_{n-1}>1$ and the equality can't happen. Therefore, all $x_k=-1$.
Assume now that $n$ is odd. Similarly as before, $s_n(s_n-s_{n-1})$ is positive, $(-1)^{n-1}s_{n-1}$ is positive again. If at least one of the $x_k<-1$ then the sum on the left is $>1$. Therefore all $x_k=-1$.
Hence $P(x)=a_n(x+1)^{n}$, or forgetting (dividing out) about the $a_n$, $P(x)=(x+1)^{m}$. This would give us $a_0=1$, $a_1=n$, $a_{n-1}=n$. We need $$1^2+n=1^2+n$$ which holds.
