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

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

• hint: consider $\mu = \min\{n \ge 1 : A^n=I\}$... Apr 9 at 20:57
• A = A^1(A^(64))^8 = A^513 = (A^27)^19 = 1
– Rafi
Apr 9 at 21:19

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 $$A = A^{19\cdot 27 - 8 \cdot 64} = (A^{27})^{19}\cdot (A^{-64})^{8} = I^{19} \cdot I^8 = I$$.
• How is $A$ invertible? Apr 9 at 21:07
• @SeanRoberson $A^{63} \cdot A = I$ Apr 9 at 21:09
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}
A "constructive" approach. $$I=A^{64}=(A^{27})^2A^{10}\Rightarrow\\I=A^{10}\Rightarrow\\I=A^{27}=(A^{10})^2A^{7}=A^7\Rightarrow\\I=A^7=A^{10}A^{-3}=A^{-3}\Rightarrow\\A^4A^{-3}=A=I$$