4
$\begingroup$

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:

$\endgroup$
  • $\begingroup$ Did you mean LCM? Also, what have you tried? $\endgroup$ – Alex R. Mar 13 '15 at 5:46
  • $\begingroup$ I want use induction $\endgroup$ – lenovo links Mar 13 '15 at 5:49
  • $\begingroup$ 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 $c \ge 1+1/k$. $\endgroup$ – marty cohen Mar 13 '15 at 6:49
1
$\begingroup$

This problem background and solution can see

Baidu tieba

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

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Good idea,maybe find recurrence relation? $\endgroup$ – lenovo links Mar 13 '15 at 7:03
  • $\begingroup$ 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). $\endgroup$ – marty cohen Mar 13 '15 at 7:12
  • $\begingroup$ You can see this problem:artofproblemsolving.com/community/c6h407542p2276540 $\endgroup$ – lenovo links Mar 13 '15 at 7:14
  • $\begingroup$ I thought of $gcd(a, b)lcm(a,b) = ab$, but wasn't able to use it. $\endgroup$ – marty cohen Mar 13 '15 at 19:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.