The Erdős–Straus conjecture states that $\frac{4}{n} = \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$, for any $n∈\mathbb{N}$ and $n\ge2$. I've done a question on this a while back, and recently I decided to keep working on it.
First, I knew that it solves for any even $n$, because $\frac{1}{n/2} + \frac{1}{n} + \frac{1}{n} = \frac{4}{n}$.
Then, with some trial and error, I figured it out for multiples of $3$: $\frac{1}{n/3} + \frac{1}{2n} + \frac{1}{2n} = \frac{4}{n}$
I eventually figured out a strategy, and here's a list up to multiples of $29$:
$$\frac{1}{n/2} + \frac{1}{n} + \frac{1}{n} = \frac{4}{n}$$
$$\frac{1}{n/3} + \frac{1}{2n} + \frac{1}{2n} = \frac{4}{n}$$
$$\frac{1}{2n/5} + \frac{1}{n} + \frac{1}{2n} = \frac{4}{n}$$
$$\frac{1}{2n/7} + \frac{1}{4n} + \frac{1}{4n} = \frac{4}{n}$$
$$\frac{1}{4n/11} + \frac{1}{n} + \frac{1}{4n} = \frac{4}{n}$$
$$\frac{1}{4n/13} + \frac{1}{2n} + \frac{1}{4n} = \frac{4}{n}$$
$$\frac{1}{6n/17} + \frac{1}{n} + \frac{1}{6n} = \frac{4}{n}$$
$$\frac{1}{6n/19} + \frac{1}{2n} + \frac{1}{3n} = \frac{4}{n}$$
$$\frac{1}{8n/23} + \frac{1}{n} + \frac{1}{8n} = \frac{4}{n}$$
$$\frac{1}{10n/29} + \frac{1}{n} + \frac{1}{10n} = \frac{4}{n}$$
I noticed that all the primes one less than a multiple of $3$ had $n$ as one of the denominators. Then I figured out an equation for where $m$ is any prime factor of $n$:
$$\frac{1}{(\frac{m+1}{3})n/m} + \frac{1}{n} + \frac{1}{(\frac{m+1}{3})n} = \frac{4}{n}$$
So now, it was easy to solve for any number with a prime factor one less than a multiple of three. And soon, I had more equations of this kind:
Where $m$ is one less than a multiple of 4:
$$\frac{1}{(\frac{m+1}{4})n/m} + \frac{1}{(\frac{m+1}{2})n} + \frac{1}{(\frac{m+1}{2})n} = \frac{4}{n}$$
Where $m$ is three less than a multiple of eight:
$$\frac{1}{(\frac{m+3}{4})n/m} + \frac{1}{(\frac{m+3}{4})n} + \frac{1}{(\frac{m+3}{8})n} = \frac{4}{n}$$
Where $m$ is three less than a multiple of twelve:
$$\frac{1}{(\frac{m+3}{4})n/m} + \frac{1}{(\frac{m+3}{6})n} + \frac{1}{(\frac{m+3}{6})n} = \frac{4}{n}$$
So now, I could easily solve for almost any number. There were just some primes, like 73, that seem to weave their way out of them. So, my question is, are there any more equations like the ones above that I can make? Also, is there anything wrong with the equations? Thank you for any answers!