# Any 'odd unit fraction' whose denominator is not $1$ can be represented as the sum of three different 'odd unit fractions'?

Let us call a fraction whose denominator is odd 'odd fraction'. Also, let us call an odd fraction whose numerator is 1 'odd unit fraction'.

Then, here is my question.

Question : Is the following true?

"Any odd unit fraction whose denominator is not $1$ can be represented as the sum of three different odd unit fractions."

Motivation : I've been asking this question. Then, I reached the above expectation.

Examples :

$$\frac 13=\frac 15+\frac 19+\frac 1{45}$$ $$\frac 15=\frac 1{7}+\frac 1{21}+\frac 1{105}$$ $$\frac 17=\frac 19+\frac 1{33}+\frac 1{693}$$ $$\frac 19=\frac 1{11}+\frac 1{51}+\frac 1{1683}$$ $$\vdots$$ $$\frac 1{99}=\frac 1{101}+\frac 1{5001}+\frac 1{16668333}$$ $$\vdots$$

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If $n$ is not a multiple of 3, then $$\frac{1}{n}=\frac{1}{n+2}+\frac{3}{(n+2)(n+4)}+\frac{1}{n(n+2)(n+4)}$$
Second try: You have the first fraction is $1/(n+2)$. Then you want to solve $$\frac{2}{n(n+2)}=\frac{1}{a}+\frac{1}{b}$$ Rearrange that into $(2a-n(n+2))(2b-n(n+2))=n^2(n+2)^2$ You can make the right-hand side equal to $pq$, where both $p$ and $q$ are $3\bmod 4$. (Take $p=n$ or $p=n+2$) Then $a$ and $b$ will be odd.
So $a=n(n+3)/2, b=n(n+2)(n+3)/2$ or $a=(n+1)(n+2)/2, b=n(n+2)(n+1)/2$
This is not correct; take any $n$ other than $7$, e.g. $n=5$ or $n=11$. In fact, $\frac1n - \frac1{n+2} - \frac3{(n+2)(n+4)} = \frac{8-n}{n(n+2)(n+4)}$ –  ShreevatsaR Oct 7 '13 at 14:31