This should be easy to prove (and well known, I bet):
Let $p$ be a prime and $M$ an $n \times n$ matrix. Then $p| \mbox{Tr} \left( M^p - M \right)$.
Little Fermat corresponds to the case where $n=1$.
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Rodrigo de Azevedo
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Hauke Reddmann
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Actually, Fermat's little theorem is when $n=1$ and when the entries of the matrix are integers. – José Carlos Santos Oct 25 '19 at 14:12
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Your bet is right, Hauke. – Dietrich Burde Oct 25 '19 at 14:27
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Integer entries, of course. BTW, does the generalization also work with the $\phi(m)$ Euler version? BTW2, https://math.stackexchange.com/questions/760848/does-fermats-little-theorem-apply-to-matrices?rq=1 – Hauke Reddmann Oct 27 '19 at 18:47
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Yes, we have the following theorem:
Theorem: Let $p$ be a prime number, $A\in M_n(\Bbb Z)$ and $k\in \Bbb N$. Then $$ p^k \mid \operatorname{tr}(A^{p^k}-A^{p^{k-1}}). $$
The case $k=1$ gives the case being asked. For a proof, see here. For Arnold's proof see here.
Dietrich Burde
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