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Let $A \in \Bbb M_3 (\Bbb Z)$ be such that $A=B^2,$ for some $B \in \Bbb M_3 (\Bbb R).$ Show that $B \in \Bbb M_3 (\Bbb Z).$

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closed as off-topic by user26857, José Carlos Santos, YiFan, uniquesolution, Song Mar 12 at 13:04

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    $\begingroup$ There have been several posts recently on the interaction between $3\times 3$ matrices over $\Bbb Z$ and over $\Bbb R$. What is going on here? $\endgroup$ – Arthur Feb 18 at 14:04
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Consider the case when $$ B = \left( \begin{matrix} \sqrt 2 & 0 & 0 \\ 0 & \sqrt 2 & 0 \\ 0 & 0 & \sqrt 2 \end{matrix} \right).$$ Then $B^2 = 2I$. It appears your claim is false.

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  • $\begingroup$ Yeah you are right. $\endgroup$ – math maniac. Feb 18 at 14:06

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