Do totally real dilogarithm ladders exist in degree $\geq 5$?

Do totally real dilogarithm ladders exist in degree $\geq 5$?

Call $\sum_j A_j \operatorname{Li}_2(x^j) + B\log^2 x \in \pi^2\mathbb{Q}, \qquad A_j, B \in \mathbb{Q},\quad x\in(0,1)\ \text{algebraic},$ a ladder, and say it is totally real if the minimal polynomial of $x$ has all roots real. Known examples:

• $n=3$: $x^3+2x^2-x-1$ (Watson, $\pi/7$), $x^3-3x-1$ (Loxton, $\pi/9$), $x^3+3x^2-1$;
• $n=4$: $x^4\mp x^3-6x^2\mp x+1$, $x^4+2x^3-7x^2+2x+1$, $x^4+4x^3-14x^2+4x+1$, over $\mathbb{Q}(\sqrt{33})$, $\mathbb{Q}(\sqrt{10})$, $\mathbb{Q}(\sqrt5)$, with Galois groups $D_4$, $V_4$, $C_4$.

Total reality is not automatic: the Gordon–McIntosh quartic $x^4+2x^3-x-1$ has only two real roots.

Question. Is there a totally real ladder of degree $n \geq 5$?

Borel's theorem makes $B(F)$ torsion for $F$ totally real, so relations exist; the issue is whether one takes the ladder shape. PSLQ to $N = 30$–$44$ on unit quintics and palindromic sextics with $f(1) = -1$, on $\mathbb{Q}(\zeta_p)^+$ for $p = 11, 13, 17$, and on Gaussian periods for $p = 29, 31, 37$, found none.

Heuristically a relation must vanish in a space of dimension $\binom{(n-1)+|S|}{2},$ where $S$ is the set of primes dividing $N_{F/\mathbb{Q}}(1-x^k)$ over the exponents used — quadratic in $n$, while the number of usable exponents grows much more slowly. But that only shows a ladder is not forced; the $n=4$ cases are not forced either.

\begin{equation} u=2\tan\tfrac{\pi}{8}\cos\tfrac{\pi}{5},\qquad v=-2\tan\tfrac{\pi}{8}\cos\tfrac{2\pi}{5}, \end{equation}

roots of $x^{4}+2x^{3}-7x^{2}+2x+1$ with Galois group $V_{4}$, generating $\mathbb{Q}(\sqrt{10})$: \begin{align} 96\operatorname{Li}_2(u)-90\operatorname{Li}_2(u^{2}) -24\operatorname{Li}_2(u^{3})+9\operatorname{Li}_2(u^{4}) +16\operatorname{Li}_2(u^{6})-2\operatorname{Li}_2(u^{12}) -12\log^{2}u &=\frac{17\pi^{2}}{6}, \label{eq:p1}\\ 96\operatorname{Li}_2(v)-90\operatorname{Li}_2(v^{2}) -24\operatorname{Li}_2(v^{3})+9\operatorname{Li}_2(v^{4}) +16\operatorname{Li}_2(v^{6})-2\operatorname{Li}_2(v^{12}) -12\log^{2}|v| &=-\frac{31\pi^{2}}{6}. \label{eq:p2} \end{align}

For $u=\tan\tfrac{3\pi}{20}$, $v=-\tan\tfrac{\pi}{20}$, roots of $x^{4}+4x^{3}-14x^{2}+4x+1$ with Galois group $C_{4}$, generating $\mathbb{Q}(\sqrt{5})$: \begin{align} 34\operatorname{Li}_2(u)-47\operatorname{Li}_2(u^{2}) +6\operatorname{Li}_2(u^{4})-2\operatorname{Li}_2(u^{5}) +\operatorname{Li}_2(u^{10})-2\log^{2}u &=\frac{2\pi^{2}}{3}, \label{eq:p3}\\ 34\operatorname{Li}_2(v)-47\operatorname{Li}_2(v^{2}) +6\operatorname{Li}_2(v^{4})-2\operatorname{Li}_2(v^{5}) +\operatorname{Li}_2(v^{10})-2\log^{2}|v| &=-\frac{4\pi^{2}}{3}. \label{eq:p4} \end{align}