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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.

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@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. –  Wadim Zudilin Jan 17 '11 at 7:14
    
@Wadim: Thanks for vote. Too easy the special case when we add condition (c) ??? –  Luis H Gallardo Jan 17 '11 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. –  Luis H Gallardo Jan 17 '11 at 8:48
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1 Answer

up vote 7 down vote accepted

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.

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@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). –  Luis H Gallardo Jan 17 '11 at 8:46
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@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. –  Laurent Berger Jan 17 '11 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 ! –  Luis H Gallardo Jan 17 '11 at 8:59
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@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. –  François Brunault Jan 17 '11 at 10:16
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@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 $\tau(N) = 0$. –  ndkrempel Jan 17 '11 at 16:06
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