I recently discovered the work done by Hanson and Nair. As I worked through Hanson's proof, an argument occurred to me regarding Legendre's Conjecture.

Clearly, my argument is wrong. It is too simple not to have been discovered in the time that these proofs have been available.

I would greatly appreciate it if someone could help me understand where my reasoning is wrong.

(1) Nair proved:

For any $n \in N^*$, we have: $$\text{lcm}({n\choose 1},2{n\choose 2},\dots,n{n\choose n})=\text{lcm}(1,2,\dots,n)$$

(2) So it follows:

$$\text{lcm}(1,2,\dots,(n^2+n)) > \left(\frac{n^2+n}{2}\right){{n^2+n}\choose{\frac{n^2+n}{2}}}$$

(3) It is straight forward to show for $n \ge 4$

$${{2n}\choose n} > \frac{4^n}{n}$$

(4) So it follows:

$$\text{lcm}(1,2,\dots,(n^2+n)) > \left(\frac{n^2+n}{2}\right)\frac{4^{n^2+n}}{n^2+n} = \frac{4^{n^2+n}}{2}$$

(5) Hanson proved:

Let $B(n)$ denote the least common multiple of the integers $1,\dots,n$. Then $$B(n) < 3^n$$

(6) So, it follows:

$$\text{lcm}(1,2,\dots,n^2) < 3^{n^2}$$

(7) Let $R(x,y) = \dfrac{\text{lcm}(1,2,\dots,x)}{\text{lcm}(1,2,\dots,y)}$

(7) So that:

$$R(n^2+n,n^2) > \frac{\frac{4^{n^2+n}}{2}}{3^{n^2}}=\frac{4^n}{2}\left(\frac{4}{3}\right)^{n^2} > 4^n$$

(8) Assume that there is no prime between $n^2$ and $n^2+n$.

(9) So, the value of $R(n^2+n,n^2)$ must be equal to the product of the relative powers of primes $p^a$ such that $n^2 < p^a \le n^2+n$.

Note: I say "relative" since for $a \ge 2$, $R(p^a,p^a-1) = p$

(10) If $p$ is a prime such that $n^2 < p^a \le (n^2+n)$, it follows that $p \le \lfloor\sqrt{n^2+n}\rfloor = n$

(11) If $n^2 < p^a \le (n^2+n)$, then $p^{a+1} = p(p^a) > pn^2$

(12) It is well known that $\prod\limits_{p \le n} p < 4^n$ where $p$ is a prime.

(13) So that, $R(n^2+n,n^2) < 4^{\pi(n)}$ where $\pi(n)$ is the prime counting function.

(14) It is well known for $n\ge8$ that $\pi(n) \le \frac{n}{2}$

(15) But then we have a contradiction since:

$$R(n^2+n,n^2) \le 4^{n/2}$$

(16) So, we reject our assumption that there is no prime between $n^2$ and $n^2+n$.


1 Answer 1


(4) So it follows:$$\text{lcm}(1,2,\dots,(n^2+n)) > \left(\frac{n^2+n}{2}\right)\frac{4^{n^2+n}}{n^2+n} = \frac{4^{n^2+n}}{2}$$

This is incorrect. It should be $$\text{lcm}(1,2,\dots,(n^2+n)) > \left(\frac{n^2+n}{2}\right)\frac{4^{(n^2+n)/2}}{(n^2+n)/2}=4^{(n^2+n)/2}$$

  • $\begingroup$ Thanks very much. That's right. That was the mistake. $\endgroup$ Commented Oct 16, 2018 at 6:03

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