Let $p$ be an odd prime number. Ramanujan's tau function satisfies:

(a) $$ \tau(p^{n+1}) = \tau(p^n)\tau(p)-p^{11}\tau(p^{n-1}) $$ for all positive integers $n>0.$ So $\tau(p)=0$ implies

(b) $$ \tau(p^{2r+1})=0, $$ and (c) $$ \tau(p^{2r})=(-1)^rp^{11r} $$ for all nonnegative integers $r \geq 0.$

Assume now that (b) happens for {\it{some}} $r \geq 0.$

Question: Can we get $\tau(p)=0$ ?

We may assume from classic Lehmer's result that $n=p^{2r+1}$ is {\it{not}} the smallest $n$ with $\tau(n)=0.$

Seems that adding condition (c) for the same $r$ works, since (essentially): if $p^k || \tau(p)$ then $k$ should be very small.