# Prove there $\operatorname{lcm}[a_{k},a_{k+1}]>ck$

Let $a_{1},a_{2},\cdots,a_{n}$ distinct postive integers .$0<c<\dfrac{3}{2}$

Show that:

there exsit infinitely postive integer $k$ such $$\operatorname{lcm}[a_{k},a_{k+1}]>ck$$

From:2015 china TsT:

• Did you mean LCM? Also, what have you tried? – Alex R. Mar 13 '15 at 5:46
• I want use induction – lenovo links Mar 13 '15 at 5:49
• You can't use induction if it is false infinitely often. For example, if, infinitely often, $a_{k}$ divides $a_{k+1}$ and $a_{k+1} \le k+1$, this is false for these $k$ for any $c \ge 1+1/k$. – marty cohen Mar 13 '15 at 6:49

## 2 Answers

This problem background and solution can see

Baidu tieba

The key is $$\dfrac{1}{[a_{i},a_{i+1}]}\le\dfrac{1}{3}\left(\dfrac{1}{a_{i}}+\dfrac{1}{a_{i+1}}\right)$$

Here's a partial solution, which shows that is it true for $c = 1$.

Want to show that $lcm[a_{k},a_{k+1}]>ck$ infinitely often.

Since the $a_k$ are distinct, $max(a_k)_{k=1}^n \ge n$.

Let $a_n$ be such that $a_n = max(a_k)_{k=1}^n$. There are an infinite number of these, since the $a_n$ are distinct positive integers. For these $a_n$, $a_n \ge n$.

Since $lcm(a, b) \ge max(a, b)$, $lcm(a_{n-1}, a_n) \ge max(a_{n-1}, a_n) \ge a_n \ge n$.

Right now, I don't see how to push $c > 1$. If we could choose an $a_{n-1}$ which does not divide one of these extreme $a_n$ infinitely often, it looks like we count get an unbounded $c$, maybe.

• Good idea,maybe find recurrence relation? – lenovo links Mar 13 '15 at 7:03
• As I said in a comment above, that relation can be false infinitely often (for example, if $a_k | a_{k+1}$ and $a_{k+1}$ is small). – marty cohen Mar 13 '15 at 7:12
• You can see this problem:artofproblemsolving.com/community/c6h407542p2276540 – lenovo links Mar 13 '15 at 7:14
• I thought of $gcd(a, b)lcm(a,b) = ab$, but wasn't able to use it. – marty cohen Mar 13 '15 at 19:40