Let $n$ be a natural number bigger or equal to two and $A,B$ two real matrices of dimension $5$ so that $A^2+B^2=\sqrt[n]{2+\sqrt[n]{2}}AB$ so $\det(AB-BA)>0$. 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.

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    $\begingroup$ How does $n$ relate to the matrices? $\endgroup$ – stuart stevenson Mar 8 '18 at 13:36
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    $\begingroup$ I think it's the numerator but the original one it s very ambiguous. It's the only way to relate. $\endgroup$ – Septimiu Cristian Mar 8 '18 at 13:38
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    $\begingroup$ What about add 2AB and consider det, or trace ? $\endgroup$ – openspace Mar 8 '18 at 15:08
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    $\begingroup$ Please develop your idea. I don't get it. $\endgroup$ – Septimiu Cristian Mar 8 '18 at 15:09

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