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I have often heard it stated that Erdős conjectured the following:

For any integer $n > 1$, there exists a prime divisor $p$ of $n$ such that
$$p > c \cdot \log \log n,$$
where $c > 0$ is a universal constant.

However, I have been unable to locate a precise reference to confirm this. Is this statement indeed true? If so, could someone provide a reference or further details about the conjecture and its context?

Thank you!

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  • $\begingroup$ So it is still $>\log \log n$? $\endgroup$
    – MAY
    Commented Nov 28 at 3:52

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The statement is false, e.g. $n=2^k$ does not have a prime factor exceeding $2$. On the other hand, if $n$ is square-free, with largest prime factor $P$, then (see here) $$\log n=\sum_{p\mid n}\log p\leq\sum_{p\leq P}\log p=\theta(P)<1.00000002P,$$ so that $P>0.99999998\log n$ by comparing the two sides. Moreover, by the Prime Number Theorem, the constant $0.99999998$ can be replaced by $1-\varepsilon$ for $n\geq n_0(\varepsilon)$ square-free; here $\varepsilon>0$ is arbitrary. ${}{}{}{}$

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  • $\begingroup$ Thank you! Do you know of any refinements or corrections to improve the lower bound for the largest prime divisor of $n $? $\endgroup$
    – MAY
    Commented Nov 28 at 4:04
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    $\begingroup$ @MAY I don't understand what you mean. I just gave you a very precise lower bound for the largest prime divisor $P$ of $n$, assuming $n$ is square-free: $P>0.99999998\log n$. If $n$ is not square-free, then there is no lower bound apart from the trivial one: $P\geq 2$. $\endgroup$
    – GH from MO
    Commented Nov 28 at 4:16
  • $\begingroup$ One more question, where is the square-freeness used in your estimation? $\endgroup$
    – MAY
    Commented Nov 28 at 4:52
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    $\begingroup$ @MAY Square-freeness is used in the equation $\log n=\sum_{p\mid n}\log p$, which is equivalent to $n=\prod_{p\mid n}p$. This equation holds if and only if $n$ is square-free. $\endgroup$
    – GH from MO
    Commented Nov 28 at 4:55

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