Solve for positive integers: $\frac{4}{13}=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$ [closed]

Solve for $x,y,z\in\mathbb{N}$

$\frac{4}{13}=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$

I tried by some general methods but they didn't help me.

Wlog. $x\le y\le z$. Then there are only finitely many cases for $x$ to consider as $\frac3x\ge \frac4{13}>\frac1x$ implies $\frac{13}4<x\le\frac{39}{4}$, i.e. $4\le x\le 9$. For each of these $x$, there are again only finitely many $y$ to consider, as we have $\frac2y\ge\frac4{13}-\frac1x>\frac1y$, i.e. $\frac1{\frac4{13}-\frac1x}<y\le \frac2{\frac4{13}-\frac1x}$. For each such $y$ check if $z=\frac1{\frac4{13}-\frac1x-\frac1y}$ is an interger (and $x\le y\le z$). So all in all this is a finite task.

I am going to try brute force. The smallest $x$ for which $\frac1x < \frac{4}{13}$ is $x=4$, so I will try that, and then I need $$\frac1y + \frac1z = \frac4{13}-\frac14 = \frac3{52}.$$

Now it is clear that $\frac3{52} = \frac1{52} + \frac2{52} = \frac1{52} +\frac1{26}$ so I am done. But if I didn't have that happy inspiration, I could repeat the process and try $y=18$ because that is the smallest $y$ for which $\frac1y \le \frac3{52}$. Then I find $$\frac1z = \frac3{52} - \frac1{18} = \frac1{468}$$ and I have a second solution.

I cannot claim this is a general method, but it did work with minimal effort and no theory.

Hint: Erdős-Strauss-Conjecture and Egyptian-Fraction-Expansion

You can get

$$\frac 4 {13} = \frac 1 4+ \frac 1 {18}+ \frac 1 {468}$$

Try using the expansion algorithm:

$$\frac{x}{y}=\frac{1}{\lceil y/x\rceil}+\underbrace{\frac{(-y)\,\bmod\, x}{y\lceil y/x\rceil}}_{*}$$

On the LHS you have your input. Then you can calculate the two terms on the RHS. The first one is the first fraction of the expansion while you have to expand the second one $(*)$ again with the same algorithm.