# Upper bound on $\text{lcm}_{p \leq x} (p^2-1)$

Consider the expression $$P(x):=\text{lcm}_{\text{p prime \leq x}} (p^2-1).$$ As the prime factors of $P(x)$ are $<x$, for every prime $q<x$ its highest power in $P(x)$ must also divide some $p^2-1$ and is therefore $\leq p^2-1 \leq x^2-1$. Thus we have an obvious upper bound of the form $P(x)<(x^2-1)^{\pi(x)}$, that is $\log P(x)=O(x)$.

Can one do better than this? What if instead of $p^2-1$ we had $p+1$ or $p-1$?

[I am thinking of substantial improvements, such as $O(x\log \log x /\log x)$ or $O(x/\log x)$. The above argument with more care leads to the upper bound $\prod_{p \leq x} (p^2-1)$, but its log is still an $O(x)$.]

• I forgot to add that with some powerful tools such as Elliott-Halberstam one can do better than this, but I need unconditional answers. Aug 2, 2014 at 10:42
• LCM of $p-1$ is tabulated at oeis.org/A058254 (but with no links or information about bounds). Similarly, LCM for $p+1$ at oeis.org/A085272 Aug 3, 2014 at 0:18

If $$q$$ is a small prime, then $$P(n)$$ will be divisible by $$q$$ for large $$n$$, because for large $$n$$ you will always be able to find a prime $$p \equiv \pm 1 \pmod{q}$$, by Dirichlet's theorem. In other words, $$P(n)$$ must be divisible by lots of small primes. However, I claim that $$P(n)$$ will not be divisible by very many large primes.
More precisely, let $$q$$ be a prime number larger than $$M := \frac{n}{\sqrt{\log(n)}}$$. Then the probability that either $$2q-1$$ or $$2q+1$$ is prime is roughly $$\frac{c}{\log(n)}$$ for some small computable constant $$c$$. And the same holds true for $$aq\pm1$$ for any $$a$$ with $$2 \le a \le \frac{n}{q} < \sqrt{\log(n)}$$. So the probability that $$q$$ divides $$P(n)$$ is upper bounded by
$$1 - \left(1 - \frac{c}{\log(n)}\right)^{\sqrt{\log(n)}} \approx 1 - e^{\frac{-c}{\sqrt{\log(n)}}} \approx \frac{c}{\sqrt{\log(n)}}$$
Therefore, ignoring factors of $$o(1)$$ for convenience,
$$P(n) \le n^{\pi(M)} n^{\frac{c(\pi(n) - \pi(M))}{\sqrt{\log(n)}}} = e^{\frac{n}{\sqrt{\log(n)}}} e^{\frac{cn}{\sqrt{\log(n)}}} = e^{\frac{(c+1)n}{\sqrt{\log(n)}}}$$