 Solution

Suppose ${\varphi }^3={\mathrm{Id}}_X$. Since $\varphi$ has no fixed points, neither does ${\varphi }^2$: if ${\varphi }^2(x)=x$, then applying $\varphi$ gives $\varphi (x)={\varphi }^3(x)=x$. The Lefschetz theorem therefore gives

For each $q$, the averaging operatoris the projection of $H_q(X;\mathbb{Q})$ onto its $⟨\varphi ⟩$-invariant subspace, so its trace is an integer. Taking the alternating sum of these traces givesThe left side is an integer. By Euler-characteristic divisibility from a fixed-point-free cyclic action,