Talk:Set of 3 Integers each Divisor of Sum of Other Two

Added conditions "coprime" and "distinct" in accordance with the analysis. --prime mover (talk) 18:22, 24 April 2020 (EDT)

Connection with unit fractions summing to 1
I wished to show a connection between the sets I have given: {1,1,1}, {1,1,2} and {1,2,3} with Sum of 3 Unit Fractions that equals 1, but I don't have anything concrete at the moment.

A similar pattern can be observed in most Sum of 4 Unit Fractions that equals 1, e.g.

for the set {21,14,6,1}, sum of 3 elements is multiple of the 4th.

They sum to a common multiple, and 1 = 1/2 + 1/3 + 1/7 + 1/42.

except for 1 = 1/3 + 1/4 + 1/4 + 1/6, which cannot be explained by this. RandomUndergrad (talk) 01:44, 25 April 2020 (EDT)


 * An interesting exploration, which goes deeper than what is needed on this page.


 * What you might want to do is to craft a new page for the purpose -- and then use that as the basis of the proof on this page (a simple corollary). Then a number of pages could be conceptually linked.


 * Bear in mind that the only reason this page exists in the first place is as a response to a single line in the Wells book, which is incomplete and inaccurate in the first place. --prime mover (talk) 04:44, 25 April 2020 (EDT)

Duh.


 * $1 = \dfrac 1 {a_1} + \dfrac 1 {a_2} + \dots + \dfrac 1 {a_n} \leadstoandfrom M = \dfrac M {a_1} + \dfrac M {a_2} + \dots + \dfrac M {a_n}$

where $M$ is the LCM of $a_i$'s. Then:
 * $\dfrac M {a_i} \in \mathbb N$


 * $\ds \frac 1 {a_k} = 1 - \sum_{\substack{1 \mathop \le j \le n \\ j \mathop \ne k} } \frac 1 {a_j}$


 * $\ds \sum_{\substack{1 \mathop \le j \le n \\ j \mathop \ne k} } \frac M {a_j} = M \paren {1 - \frac 1 {a_k}} = \frac M {a_k} \paren {a_k - 1}$

which of course is a multiple of $\dfrac M {a_k}$.

A stunningly boring conclusion after two years. Maybe I should work on that "simple corollary" before I forget.

It works with $1 = \dfrac 1 3 + \dfrac 1 4 + \dfrac 1 4 + \dfrac 1 6$ too: the set is $\set {4, 3, 3, 2}$ : $2 \divides 10, 3 \divides 9, 4 \divides 8$. --RandomUndergrad (talk) 13:09, 4 April 2022 (UTC)