Given integer n>1, prove that we can find integer k>1 and a1, a2, ......ak >1 such that

a1 + a2 +........ak = n(1/a1 + 1/a2 +....1/ak)

Does this need some complex theorem to start with? Much thanks!

Printable View

- Nov 6th 2007, 11:56 PMgentslA much needed proof
Given integer n>1, prove that we can find integer k>1 and a1, a2, ......ak >1 such that

a1 + a2 +........ak = n(1/a1 + 1/a2 +....1/ak)

Does this need some complex theorem to start with? Much thanks! - Nov 7th 2007, 09:52 AMThePerfectHacker
- Nov 8th 2007, 01:39 PMgentsl
This means n = +/- sqrt a_k therefore k >1 must apply ? smilarly for a1 + a2 +........ak = n(1/a1 + 1/a2 +....1/ak)?

- Nov 10th 2007, 10:05 AMCaptainBlack
No it means that if you set $\displaystyle a_1=a_2= .. =a_k=\sqrt{n}$ then the left hand side equals $\displaystyle k\sqrt{n}$ and the right hand side is equal to $\displaystyle n \left(\frac{k}{\sqrt{n}}\right)=k\sqrt{n} $, so the condition of the problem is satisfied by these values (at least it is when $\displaystyle n$ is a perfect square, otherwise the $\displaystyle a$'s are not integers and we still need a solution).

RonL