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.

  • $\begingroup$ @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. $\endgroup$ Jan 17, 2011 at 7:14
  • $\begingroup$ @Wadim: Thanks for vote. Too easy the special case when we add condition (c) ??? $\endgroup$ Jan 17, 2011 at 7:55
  • $\begingroup$ 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. $\endgroup$ Jan 17, 2011 at 8:48

1 Answer 1


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.

More generally, this paper should be of interest to you given your question.

  • $\begingroup$ @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). $\endgroup$ Jan 17, 2011 at 8:46
  • 2
    $\begingroup$ @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. $\endgroup$ Jan 17, 2011 at 8:51
  • $\begingroup$ @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 ! $\endgroup$ Jan 17, 2011 at 8:59
  • 5
    $\begingroup$ @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. $\endgroup$ Jan 17, 2011 at 10:16
  • 1
    $\begingroup$ @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 $\tau(N) = 0$. $\endgroup$
    – ndkrempel
    Jan 17, 2011 at 16:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.