Suppose all the zeroes of the polynomial $$p(x)=a_nx^n-a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\cdots + a_2x^2-n^2bx+b$$ in $R[x]$ are positive reals. Prove that all the zeroes are equal.
First of all, I noted that $n$ is even. Note that the sign of $a_n \equiv a_{n-2} \equiv a_2 \equiv b$ since the signs are alternating. Then, using the fact that $\displaystyle\sum_{sym} r_1r_2 \cdots r_k = \frac{a_{n-k}}{a_n}$, we get $\displaystyle\sum \frac{1}{r_i} = n^2$.
But how to proceed? I don't get any more idea.