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Reading "Find five positive integers whose reciprocals sum to 1", I wondered whether it is possible to find

Reciprocals of pairwise coprime, positive integers that sum to 1.

We can find reciprocals of arbitrary length that sum to 1 by applying the following rules to get longer sequences from shorter ones:

We have $\frac1n-\frac1{n+1}=\frac1{n(n+1)}$ and thus

$$\frac1n = \frac1{n+1} + \frac1{n(n+1)} \qquad (1)$$

and we have the trivial

$$\frac1n = \underbrace{\frac1{kn} + \frac1{kn} +\cdots}_{k \text{ times}} \qquad (2.k)$$

The coprime condition implies that all denominators must be pairwise unequal, and we can achieve that by applying the splitting rules above to get sequences of arbitrary length. For example, apply 3 times rule (2.2) and rule (1) once to get a sequence of length 5 with 5 different denominators: $$\begin{align} 1 &= \frac12 + \frac14 + \frac18 + \frac18 \\ &= \frac12 + \frac14 + \frac18 + \frac19 + \frac1{72} \end{align}$$

However, these replacement rules will always produce at least 2 denominators that are not coprime to each other.

Is there a proof that coprime, egyptian partitions of unity do not exist?

For example, the thread linked above has an answer with a computer generated list of egyptian partitions of length 5, but as far as I can see, all partitions have at least one pair with $\gcd\geqslant2$.

In the case such partitions to not exist (or only finitely many of them), then: Are there coprime, egyptian partitions for each natural number? (Excluding 1 as a denominator).

emacs drives me nuts
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1 Answers1

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Say the integers are $a_1, a_2, ..., a_n$.

$\sum_{i=1}^{n}\frac{1}{a_i}=1$

$\frac{\sum_{i=1}^{n}{\prod_{j\neq i}{a_j}}}{\prod_{i=1}^{n}{a_i}}=1$

$\sum_{i=1}^{n}{\prod_{j\neq i}{a_j}}= \prod_{i=1}^{n}{a_i}$

$\prod_{j \neq 1}{a_j}+ \sum_{i=2}^{n}{\prod_{j\neq i}{a_j}}= \prod_{i=1}^{n}{a_i} $

The first term of the left hand side of the equation is the only one not divisible by $a_1$. Therefore such sets of reciprocals do not exist.

acat3
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