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The determinant of a given square matrix $A$, with rational entries, equals 1. It is known that all entries of $A^{2015}$ are integers. Is it true that all entries of $A$ are integers?

My attemt: I've tried to construct a counter example but failed. I believe it's true but don't know how to prove it.

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Let $A = \left( \begin{array}{ccc} 1 & \frac{1}{2015} \\ 0 & 1\end{array} \right)$, then $A^{2015} = \left( \begin{array}{ccc} 1 & 1 \\ 0 & 1\end{array} \right)$.

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  • $\begingroup$ Did you use the diagonal form to compute $A^{2015}$? $\endgroup$ – prometheus21 Mar 24 '15 at 9:04
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    $\begingroup$ No; just compute $A, A^2, A^3$, then you can easily observe this. $\endgroup$ – mrka Mar 24 '15 at 9:09
  • $\begingroup$ @prometheus21 what "diagonal form" would that be? $A$ is not diagonalizable. $\endgroup$ – Omnomnomnom Mar 24 '15 at 11:24
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    $\begingroup$ @prometheus21 In general, observe $\begin{pmatrix} 1&x\\0&1 \end{pmatrix} \begin{pmatrix} 1&y \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1&x+y\\ 0&1 \end{pmatrix}$. $\endgroup$ – Kimball Mar 24 '15 at 15:53

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