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Assume $\det(A) = 1$ and all the numbers in the matrices $A^{2015}$ and $A^{2017}$ are integers. Can I say that all numbers in $A$ are integers too? How can I prove it?

Rotem ben
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    Where is this problem from? –  May 27 '15 at 07:21
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    This is a nice problem. I'm tempted to obfuscate it even further by replacing $\mathbb{Z}$ with a ring like $\mathbb{Z}/2^{64}\mathbb{Z}$ ("matrices using computer integer arithmetic") and 1 with a different unit. – user7530 May 27 '15 at 07:50
  • @user7530 Or, you could go the other way an abstract it to the point of simplicity: If a subgroup contains $g^n$ and $g^m$, then it contains $g^{(m,n)}$. – Aaron May 27 '15 at 08:05
  • I assume that by "integer matrix" you mean a matrix whose entries are all integers. Is this correct? – A.P. May 27 '15 at 09:29
  • Is this a question from an on-going contest? – JRN May 27 '15 at 12:35

2 Answers2

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Hint: If $M$ is an integer matrix with $\det(M)=1$, them $M^{-1}$ is also an integer matrix. Also, the product of two integer matrices is an integer matrix.

Aaron
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  • I didnt understand you.. may you explain please?thanks – Rotem ben May 27 '15 at 07:27
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    Do you not understand what the hint says? Or why the hint is true? Or how to use the hint? – Aaron May 27 '15 at 07:29
  • how to use it ... – Rotem ben May 27 '15 at 07:30
  • You can sort of proceed like the Euclidean algorithm. $A^{2015}$ will have determinant $1$ and have integer entries, and hence, so will $A^{-2015}$, and hence so will $A^2=A^{-2015}A^{2017}$. Is that enough to get you started? – Aaron May 27 '15 at 07:32
  • nope :/ can you help me more please? – Rotem ben May 27 '15 at 07:34
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    I'm afraid that if I say much more, it will keep you from thinking about the problem enough to actually understand what is going on. But try using the Euclidean algorithm to find the GCD of 2015 and 2017, and then look back at what I've written. – Aaron May 27 '15 at 07:36
  • the gcd of it is of course 1 ... but how can it help me? – Rotem ben May 27 '15 at 07:38
  • It seems that $A^{2013}$ is also an integer matrix. @Rotemben –  May 27 '15 at 07:40
  • why?? I dont get you. .. – Rotem ben May 27 '15 at 07:41
  • Because $A^{2013}=A^{-2}A^{2015}$. – Aaron May 27 '15 at 07:42
  • @Rotemben And the goal is $A = A^1$. –  May 27 '15 at 07:43
  • but I get like that only A^2 again ... – Rotem ben May 27 '15 at 07:44
  • Again, calculate the GCD(2015,2017) explicitly with the Euclidean algorithm (and NOT by factoring or some other method). Look at the numbers that appear and the particular calculations that you do. See if you can't find a way to transport what you do in that problem over to this problem. – Aaron May 27 '15 at 07:49
  • cant find the connection :/ – Rotem ben May 27 '15 at 07:52
  • Can you show that $A^{2013}$ is integer matrix? If yes, how about $A^{2011}$? how about $A^{2009}$? how about...... @Rotemben –  May 27 '15 at 07:53
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    so ... i get A^-2 is integer (from A^-2017 * A^2015) and then got A^2013 from A^2015*A^-2 and so on? – Rotem ben May 27 '15 at 07:58
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    Even simpler: $A^2$ is an integer matrix, so $A^{2016}=(A^2)^{1013}$ is also an integer matrix. No need for recursion, though it also works of course. Now check determinant, and apply the same trick as above to get $A^2$, and you'll find $A$. – Jean-Claude Arbaut May 27 '15 at 07:59
  • Yes, you are right. Or follow the idea of jca @Rotemben –  May 27 '15 at 08:00
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    @jca That is certainly faster, and likely something I would use in a proof I was writing for myself. I don't know that it is conceptually simpler. Sometimes calculations can uncover what is going on, but sometimes calculations can obscure the underlying idea. – Aaron May 27 '15 at 08:03
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Generally, let $\,S\,$ be a group of $\,n\times n\,$ matrices, i.e. $S$ is closed under multiplication and inverses, e.g. matrices with all integer entries having determinant $= 1$ (inverse-closed by Cramer's rule).

Theorem $\ \ A^J, A^K\in S\,\Rightarrow\, A^{\gcd(J,K)}\!\in S$

Proof $\ $ By Bezout $\ \gcd(J,K) = IJ+NK\,$ for some integers $\,I,N.$

Therefore $\ A^{\gcd(J,K)}\! = A^{IJ+NK}\! = (A^J)^I (A^K)^N \in S\ $ since $\,A^J,A^K\in S\,$ and $\,S\,$ is closed under multiplication so, by induction, under powers, including negative powers - by inverse closure (note that $\,I\,$ or $\,N\,$ may be negative in the Bezout equation).

Remark $\ $ Similarly we have $\, A^J= B^J,\, A^K = B^K\,\Rightarrow\, A^D = B^D\,$ for $\,D = \gcd(J,K)\,$ when both $\,A,B\,$ are cancellable, e.g. see the domain case here.

Bill Dubuque
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  • This was an answer to a recent question that was closed as a dupe of this question before I could post it, so I have migrated it here. – Bill Dubuque Dec 10 '16 at 16:47