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 $B \in \Bbb M_3 (\Bbb Z).$


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

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user26857, José Carlos Santos, YiFan, uniquesolution, Song
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ 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

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.

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.