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  1. every entries of $A$ is integer
  2. every entries of $A-I$ is multiple of a prime $p$ ($p\geq3$)
  3. there exists $n\ge1$ such that $A^n=I$

show that $A=I$

I tried $A=I+p^kB$ where not every entries of $B$ is multiple of $p$.

then $(I+p^kB)^n=I+np^kB+{n(n-1)\over2}p^{2k}B^2+...+p^{nk}B^n $ but how should I proceed?

thanks in advance

sacch
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  • Just an observation, but if you take your equation and reduce it mod $p^{2k}$, you get that $n$ must also be a multiple of $p^k$. I'm not entirely sure how to use this. – Aaron Jun 22 '16 at 18:53
  • A possible way to use the observation is the following result. The highest power of a prime $p$ that divides the binomial coefficient $\binom{m+n}{m}$ is equal to the number of “carries” that occur when the integers $m$ and $n$ are added in $p$-ary notation. If you can find the lowest power of $p$ in any of the terms in your sum, you would know that multiple terms have that same power. However, this seems somewhat involved, and I would like to believe there is a better way. – Aaron Jun 22 '16 at 19:17
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    The hypothesis $p\geq 3$ is indeed necessary, since for $p=2$ we have a counterexample $A=\begin{bmatrix}3&2\-4&-3\end{bmatrix}$. – Luiz Cordeiro Jun 22 '16 at 19:44

2 Answers2

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Replacing $A$ by $A^d$ for a maximal proper divisor $d$ of $n$, it suffices to prove the statement in case $n$ is a prime number.

Let us prove that $n\neq p$. By contradiction. Suppose that $n=p$. Write $A=I+p^kB$ with $B\not\equiv0\pmod p$. Then $$ I=A^p=(I+p^kB)^p=\sum_{i=0}^p\binom{p}{i}p^{ik}B^i $$ which we want to reduce modulo $p^{k+2}$. Note that the binomial coefficients $\binom{p}{i}$ are divisible by $p$ for $i=1,\ldots,p-1$. Hence $$ p^{k+2}|\binom{p}{i}p^{ik} $$ for $i=2,\ldots,p$ since $k\geq1$ and $p\geq3$. It follows that $$ I\equiv I+p^{k+1}B\pmod{p^{k+2}}, $$ fron which it follows that $B\equiv0\pmod p$. Contradiction.

Therefore, we have $n\neq p$. Since $A^n=I$, one has $(A-I)(A^{n-1}+\cdots+I)=0$. Since $A\equiv I\pmod p$, one has $A^{n-1}+\cdots+I\equiv nI\bmod p$ is invertible as a matrix with coefficients in $\mathbf F_p$. It follows that $A^{n-1}+\cdots+I$ has a nonzero determinant in $\mathbf Z$. Hence $A-I=0$, that is $A=I$.

Johannes Huisman
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We will show that $np^kB+\dots +p^{nk}B^n=0$ implies that $n=0$. To do this, assume that $n=p^lm$, where $(m, p)=1$. Then we have $mB+\dots +p^{nk-l-k}B^n=0$. We want to show that we must have every coefficient other than $m$ is a $p$-divisible integer, and then working mod $p$, we obtain a contradiction. To do this, we must show that $\binom{n}{j}p^{j-l-2}$ is an integer for $j>1$. We are studying the coefficents of $(1+px)^{p^lm}=((1+px)^{p^l})^{m}$, and an easy aurguement shows that we may reduce to $m=1$. But letting $v_p(n)$ be the maximal power of $p$ dividing $n$, we have $v_p(\binom{p^l}{j})=l-v_p(j)\ge l-j+2$, since $p>2$ and $j>1$. This is an application of Kummer's Theorem. Thus all the coefficents of the original equation vanish mod $p^{l+1}$, except the first, and we are done.

Community
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Pax
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  • you mean $p^{nk-l-k}B^n$ by $p^{nk-lk-k}B^n$? And can you please explain where did $\binom{n}{j}p^{j-l-2}$ come from – sacch Jun 22 '16 at 21:36
  • That's the coefficient of the polynomial when divided by the appropriate power of p – Pax Jun 22 '16 at 23:11
  • we divide $\binom{n}{j}p^{jk-l-k}$ by $p^{j(k-1)-k+2}$ to get $\binom{n}{j}p^{j-l-2}$? – sacch Jun 22 '16 at 23:38