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 :
$$\frac 13=\frac 15+\frac 19+\frac 1{45}$$
$$\frac 15=\frac 1{7}+\frac 1{21}+\frac 1{105}$$
$$\frac 17=\frac 19+\frac 1{33}+\frac 1{693}$$
$$\frac 19=\frac 1{11}+\frac 1{51}+\frac 1{1683}$$
$$\vdots$$
$$\frac 1{99}=\frac 1{101}+\frac 1{5001}+\frac 1{16668333}$$
$$\vdots$$
 A: Your question is essentially answered in this closely related question:
On A Splitting Equation of an Egyptian fraction to Egyptian fractions such that all produced fractions have odd denominators.
I quote verbatim from the accepted answer by MSE user Peter:

A general solution is for every positive integer $\ n\ $ :



*

*If $n$ is odd , then $$\frac{1}{3n+2}+\frac{1}{6n+3}+\frac{1}{18n^2+21n+6}=\frac{1}{2n+1}$$ is a solution with odd denominators

*If $n$ is even , then $$\frac{1}{3n+3}+\frac{1}{6n+3}+\frac{1}{6n^2+9n+3}=\frac{1}{2n+1}$$ is a solution with odd denominators

So, for every odd $\ k\ge 3\ $ we can write $\ \frac 1k\ $ with $\ 3\ $ distinct fractions with odd denominators.

Note that Peter's solution has

$$18n^2 + 21n + 6 = 3(2n + 1)(3n + 2)$$
$$6n^2 + 9n + 3 = 3(2n + 1)(n + 1).$$

A: 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 $a=(n+1)(n+2)/2, b=n(n+2)(n+1)/2$
