<s>[INST] On A Splitting Equation of an Egyptian fraction to Egyptian fractions such that all produced fractions have odd denominators. My question is: Do we have an splitting equation where we can produce fractions with odd denominators?
To split an Egyptian fraction to Egyptian fractions, we can use the splitting equation below:
$\frac{1}{n}= \frac{1}{n+1}+\frac{1}{n(n+1)}$ 
The key limitation of the above equation is the following:
If $n$ is even, then $n+1$ is odd and $n(n+1)$ is even, otherwise $n+1$ is even and $n(n+1)$ is even.
Either way, the splitting equation produces with at least one even Egyptian fraction.
An example of a splitting to Odd Egyptian fraction is given below:
$\frac{1}{3}= \frac{1}{5}+\frac{1}{9}+\frac{1}{45}$
$\frac{1}{5}= \frac{1}{9}+\frac{1}{15}+\frac{1}{45}$
$\frac{1}{7}= \frac{1}{15}+\frac{1}{21}+\frac{1}{35}$
$\frac{1}{7}= \frac{1}{9}+\frac{1}{45}+\frac{1}{105}$
$\frac{1}{9}= \frac{1}{15}+\frac{1}{35}+\frac{1}{63}$
$\frac{1}{11}= \frac{1}{21}+\frac{1}{33}+\frac{1}{77}$
The link below is useful for further details about egyptian fraction: http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fractions/egyptian.html#section9.5
 [/INST] A general solution is for every positive 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.
</s>