Let $n\ge2$ be a natural number and $A,B$ two real matrices of dimension $5$ so that $A^2+B^2=\sqrt[n]{2+\sqrt[n]{2}}AB$ and $\det(AB-BA)>0$. Prove that $n$ is a multiple of $16$.
I don't know how to solve it. I just know that the trace of $AB-BA$ is $0$ and also then, from Cayley-Hamilton, I have that $(AB-BA)^2<0$ which is false; from here I have no idea.
Questions of this kind have been asked before. For example, see this question and the various solution tricks illustrated in its answers. The following is my attempt using one of those tricks.
Let $c=\sqrt[n]{2+\sqrt[n]{2}}$ and let $z$ be a complex root to the equation $z+\frac{1}{z}=c$. Let $X=A-zB$ and $Y=A-\frac{1}{z}B$. Then \begin{align*} \frac{1}{z}XY-zYX &=\frac{1}{z}(A-zB)\left(A-\frac{1}{z}B\right)-z\left(A-\frac{1}{z}B\right)(A-zB)\\ &=\left(\frac{1}{z}-z\right)A^2-\left(\frac{1}{z^2}-z^2\right)AB+\left(\frac{1}{z}-z\right)B^2\\ &=\left(\frac{1}{z}-z\right)\left[A^2-\left(\frac{1}{z}+z\right)AB+B^2\right]\\ &=\left(\frac{1}{z}-z\right)\left(A^2-cAB+B^2\right)\\ &=0. \end{align*} Therefore $XY=z^2YX$. As $0<\det(XY-YX)=\det((1-z^2)YX)$, both $X$ and $Y$ are nonsingular. Take determinants on both sides of $XY=z^2YX$, we get $z^{10}=1$. Since $z+\frac{1}{z}=c$, the scalar $z$ is a $10$-th root of unity that satisfies the equation $\left[\left(z+\frac{1}{z}\right)^n-2\right]^n=2$, i.e., $\left[(z^2+1)^n-2z^n\right]^n=2z^{n^2}$. Now I have forgotten most of what I learned from university and do not know how to continue from here.
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