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

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?
