$A^{64}=A^{27}=I$, prove $A=I$

Question

$A^{64}=A^{27}=I$, prove $A=I$.

I tried finding the eigenvalues of A, by factoring $A^{64}-I$ but this method is too long. I'm not really sure what to do, I thought of showing $A$ is similar to $I$ but I don't know how to do that either.

Answer 1

Since $A^{64}=I$, $A$ is invertible and $(A^{-1})^{64} = I$. Also, since $64$ and $27$ are coprime, you can find a linear combination of them yielding $1$, namely $19\cdot 27 - 8 \cdot 64 = 1$.

Then $A = A^{19\cdot 27 - 8 \cdot 64} = (A^{27})^{19}\cdot (A^{-64})^{8} = I^{19} \cdot I^8 = I$.

Answer 2

A small elaboration on charmd‘s answer: we don't need to use Bezout's identity and take matrix inverses. Euclidean algorithm is perfectly sufficient: since $A^{64}=A^{27}=I$, we have $$ \begin{aligned} &A^{10}=A^{10}I^2=A^{10}(A^{27})^2=A^{64}=I\\ \Rightarrow\ &A^7=A^7I^2=A^7(A^{10})^2=A^{27}=I\\ \Rightarrow\ &A^3=A^3I=A^3A^7=A^{10}=I\\ \Rightarrow\ &A=AI^2=A(A^3)^2=A^7=I.\\ \end{aligned} $$

Answer 3

A "constructive" approach. $$I=A^{64}=(A^{27})^2A^{10}\Rightarrow\\I=A^{10}\Rightarrow\\I=A^{27}=(A^{10})^2A^{7}=A^7\Rightarrow\\I=A^7=A^{10}A^{-3}=A^{-3}\Rightarrow\\A^4A^{-3}=A=I$$