Given a positive integer $n$ , find all sequences $a_1, a_2,...., a_n$ of integers such that $\sum_{i=1}^{n} \frac{1}{a_i}=\sum_{i=1}^{n} a_i$
. I checked for small values of n like $1, 2,3,4$ .
$n=1$ , $a_1=1$ , $a_1=-1$
$n=2$ , $(1, -1), (-1, 1), (1, 1), (-1, -1)$
and $a_1=a, a_2=-a$
$n=3$
$(1, a, -a), (-1, a, -a)$ and their permutations and $(1, -1, 1), (1, -1, -1),.... $ basically those containing 1 and -1 .
A common pattern occurs for $n=4$ as well
That is either numbers occur in pairs of $(a, -a)$ or belong to $1, -1$ .
So like for $n=4$ one solution would be $(a, -a, b, -b)$ or $(a, -a, 1,-1) $ Hope you get the idea. Is there any way to prove that or even if it's not correct is there any nice to way to find all these sequences
Thanks
Edit: Also including a proof for $n=3$ similar ways also lead to proof for $n=4$ . $n=1, 2$ is trivial.
$\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}=a_1+a_2+a_3$
as if $(a_1, a_2, a_3)$ works $(-a_1, -a_2, -a_3)$ also.
So assume $a_1+a_2+a_3 \geq 0$
if $a_1,a_2,a_3 > 0$
then $a_1 \geq \frac{1}{a_1}$ equal only if $a_1=1$ . Similarly doing for other terms we get $(1, 1,1)$
If there is exactly one negative.
Let that be $a_3$
$\frac{1}{a_1}+\frac{1}{a_2}-\frac{1}{a_3}=a_1+a_2-a_3$ where $a_i$ is in N.
wel$LHS<\frac{1}{a_1}+\frac{1}{a_2}<=2$
So it must be 1 or 0.If its 1 , we can using inequality get a1=1, a2=b or reverse. If its 0 then just substitute $a_3=a_1+a_2$ .
If there are two negatives things are similar as well