# Estimate on radical of $2^n \pm 1$

Not sure if this belongs to MO or not. Are there any lower bound on radical of $2^n \pm 1$?

We recall that radical of an integer $rad(k)$ is a product of primes which divide $k$.

As an example, if the abc-conjecture is true in the form $max(|a|,|b|,|c|) \leq rad(abc) ^2$ then $$rad(2^n \pm 1) \geq 2^{n/2 - 1}.$$ I wonder if this estimate is proven (or perhaps conjectured) by anyone? Are there any nontrivial results here?

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The inequalities in the comment seem to be intended in the other direction. – user9072 Jan 15 '13 at 18:15
$2^n + 1 = (2^n + 1) \rightarrow 2^n+1 \leq rad (1\times 2\times (2^n+1))^2 \rightarrow 2rad(2^n+1) \geq \sqrt{2^n+1}$ Thank you, quid, sorry for that – Ostap Chervak Jan 15 '13 at 18:23
Yyou could use the Cunningham project for numerical evidence to help decide what the conjecture should be. Gerhard "Ask Me About System Design" Paseman, 2013.01.15 – Gerhard Paseman Jan 15 '13 at 18:34
There are certainly nontrivial results, e.g., from work of Stewart and Yu towards the ABC conjecture it seems one gets $\mathrm{rad}(2^n\pm 1) \geq n^{3-o(1)}$ as $n\to\infty$. – Anonymous Jan 16 '13 at 17:33
According to 'On the largest prime factor of the Mersenne Numbers', written by Ford, Luca, and Shparlinski, Schinzel proves in 'On primitive prime factors of $a^n - b^n$' that the largest prime factor of $2^n-1$ is at least $2n-1$ for all $n \ge 13$. – Woett Jan 17 '13 at 0:32

## 1 Answer

The paper arXiv:1409.2974v1 proves a lemma regarding $2^L-1$.

Let $n$ be a positive integer, and define $L = \text{lcm}[1, 2, \dots n]$ and $t = \lfloor \frac{\log(n)}{\log(2)} \rfloor$. Then $L*\text{rad}(2^L-1) \le 2^t(2^L-1)$.

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