Tartaglia and complex solutions to the system $a^n+b^n+c^n+n\left(\frac1a+\frac1b+\frac1c\right)=3(n+1)$ and others?

Question

In 1535, there was a challenge between Tartaglia and Antonio Fiore to solve certain cubics, one of which was (OEIS link),

$$r^3-3r-10=0$$

This has a nice solution, connected to the fundamental unit $U_6 = 5+2\sqrt6\,$ since,

$$r=\big(5+2\sqrt6\big)^{1/3}+\big(5-2\sqrt6\big)^{1/3} = 2.612887\dots$$

Note that $(p,q)= (5,2)$ solve the Pell equation $p^2-6q^2=1$.

---

I. Cubic

Fast forward almost 500 years to a 2023 MSE post by TheSimpliFire which seeks to solve the system,

\begin{align}a\,+b\,+c\,+\left(\frac1a+\frac1b+\frac1c\right)&=6\\ a^3+b^3+c^3+3\left(\frac1a+\frac1b+\frac1c\right)&=12\\ a^5+b^5+c^5+5\left(\frac1a+\frac1b+\frac1c\right)&=18\end{align}

This has real solution $(a,b,c)=(1,1,1)$. However, if we allow complex solutions, then Tartaglia's cubic suddenly appears. The roots of the "cubic",

$$5x^3 + 5(r + 2)x^2 + (r + 8)r x - r = 0$$

for each root $r_k\,$ of $\,r^3-3r-10=0$ yield a new triple $(x_1, x_2, x_3) = (a,b,c)$. For example, let $r_1\approx2.612887$, then the three roots of the cubic in $x$ yield a solution to the system above,

$$(a,b,c) \approx (0.087705,\, -2.35029 + 0.65911i,\, -2.35029 - 0.65911i)$$

If we use $r_2 \approx -1.306443 + 1.456154i$, then it yields a second triple and $r_3 \approx -1.306443 - 1.456154i$ yields a third, so the three triples are roots of a $9$-deg equation with integer coefficients (WolframAlpha link),

$$-2 + 48 y - 354 y^2 + 669 y^3 + 1032 y^4 + 147 y^5 + 30 y^6 + 255 y^7 + 150 y^8 + 25 y^9 = 0$$

---

II. Quartic

The analogous system,

$$a^n+b^n+c^n+d^n+n\left(\frac1a+\frac1b+\frac1c+\frac1d\right)=4(n+1)$$

for $n=(1,3,5,7)$ has solution $a=b=c=d=1$ but can also be solved by the complex roots of a "quartic" whose coefficients are determined by the sextic,

$$r^6 - 15 r^4 - 60 r^3 - 144 r - 720 = 0$$

It has nice discriminant $D=2^{24}\times 3^{16}\times 5^6\times 7$. Unfortunately, the sextic does not have a solvable Galois group. The quadruple $(a,b,c,d)$ are then roots of a $4\times6=24$-deg equation.

---

III. Quintic

Likewise,

$$a^n+b^n+c^n+d^n+e^n+n\left(\frac1a+\frac1b+\frac1c+\frac1d+\frac1e\right)=5(n+1)$$

for $n=(1,3,5,7,9)$. My guess is the complex solutions $(a,b,c,d,e)$ are roots of a "quintic" with coefficients determined by a decic. Hence, they are roots of a $5\times10=50$-deg equation.

---

IV. Question

Q: Is it true that the sequence of degrees is $\dfrac{(m-1)m^2}2=0,2,9,24,50,\dots$? (A006002)