Let $A^{27}=A^{64}=I$, show that $A=I$ Let $A$ be a square matrix, $A^{27}=A^{64}=I$,
show that $A=I$

Use the fact that $A^{x} * A^{y}=A^{x+y}$
$A^{27}=I$ implies $A^{54}=I$.
$A^{54}=I$ and $A^{64}=I$ imply $A^{10}=I$.
$A^{10}=1$ implies  $A^{30}=I$
$A^{30}=I$ and $A^{27}=I$ imply $A^{3}=I$.
$A^{3}=1$ implies  $A^{9}=I$
$A^{9}=I$ and $A^{10}=I$ imply $A^{1}=A=I$.
This is essentially applying the euclidean algorithm to find the gcd of 27 and 64.
