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PROVED This has been solved in the affirmative.
If $P(m)$ is the greatest prime divisor of $m$, then is it true that\[\frac{P(2^n-1)}{n}\to \infty\]as $n\to \infty$?
Schinzel [Sc62] proved that $P(2^n-1)>2n$ for $n>12$. In [Er65b] Erdős also asks about $P(n!+1)$.

Stewart [St74b] proved that this conjecture is true if we restrict $n$ to those integers with $<\frac{1}{\log 2}\log\log n$ many prime factors.

This was proved in the affirmative by Stewart [St13], who proved that\[P(2^n-1)\gg n^{1+\frac{1}{104\log\log n}}\]for all large $n$.

The case of $P(n!+1)$ appears to be open still. Murty and Wong [MuWo02] proved that\[P(n!+1)>(1+o(1))n\log n\]assuming the abc conjecture. The best-known unconditional result, due to Lai [Li21], is that\[\limsup \frac{P(n!+1)}{n} \geq 1+9\log 2\approx 7.238.\]

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This page was last edited 13 October 2025.

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Additional thanks to: Alfaiz and Mehtaab Sawhney

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T. F. Bloom, Erdős Problem #977, https://www.erdosproblems.com/977, accessed 2026-01-16