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Question: Is there an Egyptian fraction representation for $1$ where all the fractions have odd denominators?

I tried to generate one below:

$$\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+\frac{1}{11}+\frac{1}{13}+\frac{1}{23}+\frac{1}{721}+\frac{109}{106711605}.$$

The last term can be further decomposed to: $$\frac{1}{979007}+\frac{158}{1.04471\cdot 10^{14}}.$$

or, it is impossible for any collection of $\frac{1}{n}$ where $n$ is odd to produce $1$?

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  • $\begingroup$ I guess the denominators should be distinct , right ? $\endgroup$
    – Peter
    Commented Mar 17, 2020 at 14:16
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    $\begingroup$ yes, they should be distinct.@Peter $\endgroup$
    – Keneth Adrian Dagal
    Commented Mar 17, 2020 at 14:19

3 Answers 3

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The answer is yes and I have shown how, but I can't find the previous question. It takes at least $9$ fractions. This page has many expansions. The one with the smallest maximum denominator is $$1=\frac 13+\frac 1 5+\frac 1 7+\frac 1 9+\frac 1 {11}+\frac 1{ 15}+\frac 1 {35}+\frac 1{45}+\frac 1{ 231}$$

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Technique: Find an odd abundant number--more precisely, an odd semi-perfect number: ($945$ is the smallest).

Write it as a sum of its factors: $945=315+189+135+105+63+45+35+27+21+7+3$.

Then divide this equation by the original number: $945$, keeping the terms on the right separated:

$1=\frac13+\frac15+\frac17+\frac19+\frac{1}{15}+\frac{1}{21}+\frac{1}{27}+\frac{1}{35}+\frac{1}{45}+\frac{1}{135}+\frac{1}{315}$

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$$\frac 13+\frac 15+\frac 17+\frac 19+\frac{1}{15}+\frac{1}{19}+\frac{1}{21}+\frac{1}{25}+\frac{1}{173}+\frac{1}{1294257}+\frac{1}{2233466948475}$$ is a solution

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