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Given a polynomial $f$, it is known believed that the number of smooth values of $f$ has a positive proportion (for fixed $u$, $\lim_{X\rightarrow\infty} \frac{|\{ n < X\ :\ f(n)\ is\ X^u\ smooth \}|}{X} > 0$ (I think this formulation is fine, please correct if not)).

What is known believed if we only go over (positive) prime numbers, i.e. $f(p)$?

This question came to me while looking over a paper uploaded to arXiv, that claims to prove Lehmer's conjecture - that $\tau(n)\not=0$, $\tau$ being Ramanujan's function. Lemma 2 of the paper says that if $p=-1\pmod{691}$, then $A(p):=(65(1+p^{11})+691(1+p^5))/252$ has a prime factor greater than $p$. So if $p=691k-1$, then $A(p)=kg(k)/252$ where $$g(k) = 1613220864834440404965591741065k^{10} - 25680795243384724246919694865k^9 + 185823409865301912061647575k^8 - 806758653539660978559975k^7 + 2335046754094532499450k^6 - 4730919617557663530k^5 + 7004022584795281k^4 - 8217140034755k^3 + 8420377435k^2 - 7245135k + 4170$$

Just in case you're wondering, $g(k)$ is always divisible by 252 (and also by 691). But it is irreducible. The claim in Lemma 2 of the paper essentially says that $g(k)$ cannot be $691k$ smooth when $691k-1$ is prime. The proof in the article is long and hard to read, and I realised this question is independently of interest.

Here's a link to the article: http://arxiv.org/abs/1406.3607

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    $\begingroup$ I don't think we know that values of polynomials get smooth (except for $u$ in some limited range), let alone for polynomials evaluated at primes. $\endgroup$
    – Lucia
    Commented Nov 2, 2014 at 19:46
  • $\begingroup$ I overshot there a bit... :) $\endgroup$
    – Dror Speiser
    Commented Nov 2, 2014 at 20:07
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    $\begingroup$ Standard heuristics would suggest that this proportion should be $\rho(d/u)$, where $d$ is the degree of $f$ and $\rho$ is Dickman's function. So, the claimed lemma should fail about $\rho(11) \approx 10^{-12}$ of the time, suggesting that counterexamples exist but would require an immense amount of computation to locate numerically. $\endgroup$
    – Terry Tao
    Commented Nov 2, 2014 at 21:06
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    $\begingroup$ For what it's worth: there is at least one error in the proof of Lemma 2 in the arXiv preprint, namely in the third line of (21). An equality of integers is claimed there due to (18), but (18) only gives the claimed equality modulo $p^{i-1}(p-1)$. This is insufficient to then deduce the next line of (21). $\endgroup$
    – Terry Tao
    Commented Nov 2, 2014 at 21:48

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