Let $A^{27}=A^{64}=I$, show that $A=I$ Let $A$ be a square matrix, $A^{27}=A^{64}=I$,
show that $A=I$
 A: Hint: 27 and 64 are coprime and consider Bezout's identity.
A: Use the Jordan normal form and you have
$$ (D+N)^{27}= (D+N)^{64}$$ 
Hence $N=0$ 
As they are coprimes they can't be root of unities. Hence $D=I$.
A: 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.
A: Because $\gcd(27, 64) = 1$, there are integers $m, n$ such that $27m = 64n + 1$.
We know that
$$A^{27m} = \left(A^{27}\right)^m = I^m = I$$
but at the same time, we have
$$A^{27m} = A^{64n + 1} = AA^{64n} = A\left(A^{64}\right)^n = AI^n = A$$
Thus, $A = I$.
A: Since $A^{27}-I=0$ and $A^{64}-I=0$, the minimal polynomial $m_A$ of $A$ divides both $x^{27}-1$ and $x^{64}-1$. Since 27 and 64 are co-prime, the greatest common divisor of these two polynomials is $x-1$ and thus $A-I=0$ or $A=I$.
A: First of all, since $A^{27}=I$, $A$ is invertible.
Since $19\cdot27-8\cdot64=1$, we have
$$
A^1=\left(A^{27}\right)^{19}\left(A^{64}\right)^{-8}=I^{19}I^{-8}=I
$$
