Given an irreducible solvable equation $P(x)=0$ of prime degree $p>2$ with rational coefficients and $\zeta^p=1$, define the usual *Lagrange resolvents* of the roots $x_i$ as,

$$R_n = \big(x_1+x_2\zeta^n+x_3\zeta^{2n}+\dots+x_p\zeta^{(p-1)n}\big)^p$$

where the $R_n$ are the roots of an equation of deg $p-1$ with rational coefficients.

* Question*: For any prime $p>2$, can half of the resolvents be $R_n = 0$ while the other half yields an

*irreducible*equation of deg $\frac{p-1}{2}$?

For example, the case $p=3$ is trivial, the *DeMoivre quintic* is for $p=5$, while the septic quadrinomial,

$$7x^7+14x^4+7x^3-1=0\tag{1}$$

can be solved as,

$$x = 7^{-2/7}\,\big(y_1^{1/7}+y_2^{1/7}+\dots+y_6^{1/7}\big)=0.423604\dots$$

and the $y_i$ are the roots of the "*sextic*",

$$y^3(7y^3-7y+1)=0\tag{2}$$

If it can be done for $p=11$ and higher, can one give an example? (P.S. The cubic in (2) looks so familiar.)