This post is inspired by this earlier question:
I quote:
An example of a splitting to Odd Egyptian fraction is given below: $$\frac{1}{3}= \frac{1}{5}+\frac{1}{9}+\frac{1}{45}$$ $$\frac{1}{5}= \frac{1}{9}+\frac{1}{15}+\frac{1}{45}$$ $$\frac{1}{7}= \frac{1}{15}+\frac{1}{21}+\frac{1}{35}$$ $$\frac{1}{7}= \frac{1}{9}+\frac{1}{45}+\frac{1}{105}$$ $$\frac{1}{9}= \frac{1}{15}+\frac{1}{35}+\frac{1}{63}$$ $$\frac{1}{11}= \frac{1}{21}+\frac{1}{33}+\frac{1}{77}$$ Notice that $$3 \mid 45$$ $$5 \mid 45$$ $$7 \mid 35$$ $$7 \mid 105$$ $$9 \mid 63$$ $$11 \mid 77$$
My question is:
If $$\frac{1}{y}=\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}$$ where all of $y$, $x_1$, $x_2$, and $x_3$ are distinct positive integers, and $x_1 < x_2 < x_3$, does it follow that $y \mid x_3$?
MY ATTEMPT
In the accepted answer, we have the following:
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.
Notice that
$$18n^2 + 21n + 6 = 3(2n + 1)(3n + 2)$$ $$6n^2 + 9n + 3 = 3(2n + 1)(n + 1).$$
Follow-Up Questions
If the answer to my first question is YES, how can such a claim be proved? Lastly, if the answer to my first question is NO, what is/are some of the smallest counterexample(s)?
The link below is useful for further details about Egyptian fractions: http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fractions/egyptian.html#section9.5
