# If Ramanujan's tau function has a prime power zero then $\ldots$

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.

• @Luis, I vote down since I see no mathematical content. The famous conjecture $\tau(p)\ne0$ is plausible enough (maybe, even more than RH, at least to me). I vote to close as this is not a research level. Jan 17, 2011 at 7:14
• @Wadim: Thanks for vote. Too easy the special case when we add condition (c) ??? Jan 17, 2011 at 7:55
• The question is \it{NOT} about discovering a possible zero of $\tau$ (out of reach of course) It is modestly to see if some necessary conditions may be sufficient or not. Jan 17, 2011 at 8:48

Theorem 2 of Lehmer's paper "The vanishing of Ramanujan's function $\tau(n)$" says that the smallest $n$ for which $\tau(n)=0$ is prime.
• @Laurent: I know the paper. Using it I putted in the question that we may assume that the power of $p$ (assumed to be a possible zero of $\tau$) is not the smallest one. Indeed, Lehmer proved these result by playing with the known congruences for $\tau$ together with some computations. (He succedded since being the smallest possible zero implies reasonable computations). Jan 17, 2011 at 8:46
• @Luis: if I'm not mistaken, the proof of theorem 2 consists in showing that if $\tau(p^n)=0$ then $\tau(p)=0$ which does answer your question. Jan 17, 2011 at 8:51
• @Laurent: Sure, but assuming that $N=p^n$ is the "smallest" $N$ such that $\tau(N)=0.$ If we do not assume this the proof of Lehmer vanishes ! Jan 17, 2011 at 8:59
• @Luis, condition (c) also implies $\tau(p)=0$ : let $1-\tau(p)X+p^{11} X^2 = (1-\alpha X)(1-\beta X)$ with $|\alpha|=|\beta| = p^{11/2}$ (by Deligne). Condition (c) says that $\sum_{i=0}^{2r} \alpha^i \beta^{2r-i} = (-1)^r p^{11r}$ which implies that $\alpha/\beta$ is a root of unity, so some $\tau(p^k)$ is zero. Jan 17, 2011 at 10:16
• @GH: I think you misunderstood - Luis was concerned that the proof would only work for N being the smallest prime power with $\tau(N) = 0$, as opposed to larger prime powers for which $\tau(N) = 0$. Jan 17, 2011 at 16:06