If A, B are two integer square matrices of the same size such that $A\equiv B\pmod n$, is $A^p\equiv B^p \pmod{pn} $ for a prime p dividing n?
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1Just to make sure, the congruence of the matrix is defined componentwise, right? – YiFan Tey Oct 14 '21 at 01:23
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2Are you sure that this even holds for $1 \times 1$ matrices? – Ben Grossmann Oct 14 '21 at 01:39
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@YiFan yes. And sorry, I meant to assume that p divides n. I think it holds for 1x1 matrices in that case – Asvin Oct 14 '21 at 01:42
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Is $n$ arbitrary? – Kenta S Oct 14 '21 at 01:42
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@TorstenSchoeneberg But $2\not\mid3$... – Kenta S Oct 14 '21 at 01:47
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@KentaS Arbitrary with only the assumption that $p\mid n$. – Asvin Oct 14 '21 at 01:51
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let $n=p=2$
let $A=\begin{pmatrix}0&0\\1&0\end{pmatrix}$ and $B=\begin{pmatrix}0&2\\1&0\end{pmatrix}$
$A^2$ is the zero matrix and $B^2=\begin{pmatrix}2&0\\0&2\end{pmatrix}$
$A\equiv B\pmod{2}$ but $A^2\not\equiv B^2\pmod{4}$
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