prove if we want that the sum of some fractions be $1$ and the denominators of one of them is $d$ then another denominators should divisible by $d$ or $d$ should be divisible to another denominators.
It seems to be easy I tried to prove it.I first tried some cases.
$1=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}$
Here we can see that $6$ is divisible by $3$. Also here $6$ is divisible by $2$ .But I want to prove one of the denominators but here two of them is possible.After trying a lot I cannot found any proofs.Any hints?
update1: the numerator should be prime
As stated, the claim offered for proof is not true. For example: $$\frac {7}{12} +\frac{4}{15} +\frac{3}{20} = 1$$
An alternative, which avoids composite numerators:$$\frac {1}{12} +\frac{13}{15} +\frac{1}{20} = 1$$