# A much needed proof

• November 6th 2007, 11:56 PM
gentsl
A 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!
• November 7th 2007, 09:52 AM
ThePerfectHacker
Quote:

Originally Posted by gentsl
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!

Say a_k = x then on LHS we have k*x on RHS we have n*k/x.
So we require that x^2 = n so chose x = sqrt(n) for all a_1,...,a_k
• November 8th 2007, 01:39 PM
gentsl
This means n = +/- sqrt a_k therefore k >1 must apply ? smilarly for a1 + a2 +........ak = n(1/a1 + 1/a2 +....1/ak)?
• November 10th 2007, 10:05 AM
CaptainBlack
Quote:

Originally Posted by gentsl
This means n = +/- sqrt a_k therefore k >1 must apply ? smilarly for a1 + a2 +........ak = n(1/a1 + 1/a2 +....1/ak)?

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

RonL