# Is it possible to scale roots such that half of the coefficients of two real monic polynomials agree?

Suppose we have two real monic polynomials \begin{align*} & p_1(t) = t^n + a_{n-1} t^{n-1} + \dots + a_0, \\ & p_2(t) = t^n + b_{n-1} t^{n-1} + \dots + b_0. \end{align*} Let $\alpha = (\alpha_1, \dots, \alpha_n) \subset \mathbb C$ and $\beta = (\beta_1, \dots, \beta_n) \subset \mathbb C$ be the roots of $p_1$ and $p_2$ respectively and assume no root in $\alpha$ and $\beta$ equal $0$. Necessarily, $\alpha$ and $\beta$ must be invariant under conjugation since the coefficients are real.

Does there exist nonzero real numbers $r_1, r_2$ such that the monic polynomials $\hat{p}_1(t)$ with roots $r_1 \alpha= (r_1 \alpha_1, \dots, r_1 \alpha_n)$ and $\hat{p}_2(t)$ with roots $r_2 \beta = (r_2 \beta_1, \dots, r_2 \beta_2)$ have either $a_{2j} = b_{2j}$ for $j =0, 1, \dots, [\frac{n}{2}]$ or $a_{2j+1} = b_{2j+1}$ for $j = 0, 1, \dots, [\frac{n-1}{2}]$ where $[s]$ denotes the greatest integer less than $s \in \mathbb R$?

Quadratic case seems always true. Is there a way to see the existence or nonexistence for higher orders?

It is not possible in general for $\,n \gt 2\,.$ By Vieta's relations:
$$\hat{p}_1(t)=t^n + r_1 a_{n-1}t^{n-1}+r_1^2a_{n-2}t^{n-2}+\ldots +r_1^{n-1}a_1t + r_1^na_0 \\ \hat{p}_2(t)=t^n + r_2 b_{n-1}t^{n-1}+r_2^2b_{n-2}t^{n-2}+\ldots +r_2^{n-1}b_1t + r_2^nb_0 \\$$
For two coefficients to end up being equal the condition is $\,r_1^ia_{n-i}=r_2^ib_{n-i}\,$. Another pair of coefficients can become equal iff $\,r_1^ja_{n-j}=r_2^jb_{n-j}\,$. But (assuming non-zero coefficients) the two equations are consistent only iff $\,\left(\dfrac{b_{n-i}}{a_{n-i}}\right)^j=\left(\dfrac{b_{n-j}}{a_{n-j}}\right)^i=\left(\dfrac{r_1}{r_2}\right)^{ij}\,$. Therefore, it is not possible to match more than one pair of coefficients with this construct in general, so $\,n \le 3\,$.
The case $\,n=3\,$ is excluded because the coefficients of the linear terms can be made to match $\,r_1^2a_1=r_2^2b_1\,$ only if $\,a_1\,$ and $\,b_1\,$ have the same sign.