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If $A$ is a square matrix such that $A^{27}=A^{64}=I$ then $A=I$.

What I did is to subtract I from both sides of the equation:

$$A^{27}-I=A^{64}-I=0$$

then:

\begin{align*} A^{27}-I &= (A-I)(A+A^2+A^3+\dots+A^{26})=0\\ A^{64}-I &= (A-I)(A+A^2+A^3+\dots+A^{63})=0. \end{align*}

So from what I understand, either $A=I$ (as needed) or $A+A^2+A^3+\dots+A^{26}=0$ or $A+A^2+A^3+\dots+A^{63}=0$.

At this point I got stuck. By the way, I found out that $A$ is an invertible matrix because if $A^{27}=I$ then also $A^{26}A=AA^{26}=I$ then $A^{26}=A^{-1}$.

Also I thought to use the contradiction proving by assuming that $A+A^2+A^3+\dots+A^{63}=0$, but because $A^{27}=I$, then: $$A+A^2+A^3+\dots+A^{26}+I+A^{28}+\dots+A^{53}+I+A^{55}+\dots+A^{63}=0$$ but yet nothing.

Would appreciate your guidance, thanks!

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    $\begingroup$ Note, if the product of two matrices is zero, it is not necessarily the case that one of the matrices is zero. That is, $AB = 0 \nRightarrow A = 0$ or $B = 0$. $\endgroup$ – Michael Albanese Mar 20 '16 at 18:18
  • $\begingroup$ @MichaelAlbanese good point. wondering why I didn't notice. thanks $\endgroup$ – Ami Gold Mar 20 '16 at 18:20
  • $\begingroup$ Hint: Compute $$\left(A^{64}\right)^8\cdot A=A^{64\times8+1}=A^{513}=A^{27\times19}=\left(A^{27}\right)^{19}.$$ $\endgroup$ – Did Mar 20 '16 at 18:41
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\begin{align*} & I = A^{64} = A^{2(27) + 10} = (A^{27})^2A^{10} = A^{10}\\ \implies & I = A^{27} = A^{2(10) + 7} = (A^{10})^2A^7 = A^7\\ \implies & I = A^{10} = A^{1(7)+3} = (A^7)^1A^3 = A^3\\ \implies & I = A^7 = A^{2(3) + 1} = (A^3)^2A = A. \end{align*}

This is nothing more than the Euclidean algorithm applied to the exponents. The same procedure can be used to show that if $A^p = I$ and $A^q = I$ with $p$ and $q$ coprime, then $A = I$.

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  • $\begingroup$ wow this is tricky! liked your solution, however I wonder I could think about it all by myself. what was your motivation for such a solution? $\endgroup$ – Ami Gold Mar 20 '16 at 18:30
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    $\begingroup$ Well, I knew I could write $A^{64}$ as $A^{64-27}A^{27} = A^{37}$, so $A^{37} = I$. But then I could just keep repeating this trick, each time getting a smaller power of $A$ which was the identity, until that power was $1$. $\endgroup$ – Michael Albanese Mar 20 '16 at 18:34
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Firstly, since $A A^{26} = A^{26}A=I$, then $A$ is an invertible matrix.

Use the fact that $\gcd(27,64)=1$: hence there exist some $a,b \in \Bbb{Z}$ such that $1=27a+64b$. Now, compute $$A=A^1=A^{27a+64b}=(A^{27})^a(A^{64})^b=I^aI^b=I$$

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  • $\begingroup$ Nice. This is much quicker than the method I suggested. $\endgroup$ – Michael Albanese Mar 20 '16 at 18:19
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Hint: if matrix $A$ has eigenpair $(\lambda, v )$, then $A^{27}$ and $A^{64}$ have eigenpairs $(\lambda^{27}, v)$ and $(\lambda^{64}, v)$. At the same time $A^{27} v = A^{64} v = 1 \cdot v$. Could you proceed from here?

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