# On A Splitting Equation of an Egyptian fraction to Egyptian fractions such that all produced fractions have odd denominators - Part II

This post is inspired by this earlier question:

On A Splitting Equation of an Egyptian fraction to Egyptian fractions such that all produced fractions have odd denominators.

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

• Thank you for providing the two counterexamples in your last comment, @quasi! If you write that out as an actual answer, then I will accept it. =) Apr 11, 2020 at 5:27

A counterexample with $$y=2$$: $$\frac{1}{2}=\frac{1}{3}+\frac{1}{10}+\frac{1}{15}$$ A counterexample with all odd denominators: $$\frac{1}{9}=\frac{1}{11}+\frac{1}{63}+\frac{1}{231}$$ A counterexample such that none of $$y,x_1,x_2,x_3$$ is a multiple of one of the others: $$\frac{1}{10}=\frac{1}{15}+\frac{1}{55}+\frac{1}{66}$$