I wanted to list all the natural number solutions $(d_1,d_2,...,d_n)$ to the equation:
$$\sum_1^n \frac1{d_i} = 1$$
I could not succeed. I noted that for $n=4$, $(2,4,8,8), (3,3,6,6), (2,3,12,12), (2,6,6,6),$ and $(4,4,4,4)$ are the solutions I could find.
It seems to me that if a solution $\underline{\bf d}$ is such that g.c.d $(d_i, d_j) > 1$ for all $i,j$, then either $d_i | d_j$ or $d_j | d_i$ for all $i,j$.
For example, apart from $(2,3,12,12)$ which does not qualify, all other solutions satisfy my guess. So I was wondering if the guess is true in general.
To be precise:
Given $n$ naturals $(d_1,d_2,...,d_n)$ such that
$\displaystyle \sum_1^n \frac1{d_i} = 1.$
$\text{g.c.d}(d_i,d_j) > 1$, for all $1\leq i,j \leq n$.
Then prove or disprove that either $d_i | d_j$ or $d_j | d_i$ for all $i,j$.
Thanks!
$$ \frac{1}{4} +\frac{1}{4} + \frac{1}{6} + \frac{1}{6} + \frac{1}{12} + \frac{1}{12} = 1 $$