<s>[INST] Solving an equation in $\mathbb N$ I am trying to solve the equation $$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{3}{5}.$$
I have made the following progress:
1) $x, y z$ have to be larger than $1$
2) only one of x, y, z can be $2$; rest should be larger
3) WLOG I have assummed $2\leqslant x\leqslant y\leqslant z$.  Knowing this, $x$ has to be smaller than $5$. How to find $y$ and $z$ for all cases of $x \in\{2, 3, 4, 5\}$ to be sure that I am finding all cases?
Thanks.
 [/INST] This is not an exhaustive list. But it is at least a partial solution.
$$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{3}{5}.$$
We know that $\dfrac{1}{ab}=\dfrac{1}{a(a+b)} + \dfrac{1}{b(a+b)}$
Let $x=2$ and you get $\frac{1}{y}+\frac{1}{z}=\frac{1}{10}.$
We can use $a=2$ and $b=5$ to get
$$\frac{3}{5} = \frac{1}{2}+\frac{1}{14}+\frac{1}{35}.$$
Let $x=3$ and you get $\frac{1}{y}+\frac{1}{z}=\frac{4}{15}.$
We consider $\gcd(a,b)=1$. Then 
$\dfrac{1}{ag}+\dfrac{1}{bg}=\dfrac{a+b}{abg}=\dfrac{4}{15}$.
It seems $a=1$, $b=3$, and $g=5$ will work. We get
$$\frac{3}{5} = \frac{1}{3}+\frac{1}{5}+\frac{1}{15}.$$
Let $x=4$ and you get $\frac{1}{y}+\frac{1}{z}=\frac{7}{20}.$
We consider $\gcd(a,b)=1$. Then 
$\dfrac{1}{ag}+\dfrac{1}{bg}=\dfrac{a+b}{abg}=\dfrac{7}{20}$.
No solution there. Next we try
$\dfrac{1}{ag}+\dfrac{1}{bg}=\dfrac{a+b}{abg}=\dfrac{14}{40}$.
Then $(a,b,g)=(4,10,1)$ gives us
$$\frac{3}{5} = \frac{1}{4}+\frac{1}{4}+\frac{1}{10}.$$
Let $x=5$ and you get $\frac{1}{y}+\frac{1}{z}=\frac{2}{5}.$
We consider $\gcd(a,b)=1$. Then 
$\dfrac{1}{ag}+\dfrac{1}{bg}=\dfrac{a+b}{abg}=\dfrac{2}{5}$.
Then $(a,b,g)=(1,1,5)$ gives us
$$\frac{3}{5} = \frac{1}{5}+\frac{1}{5}+\frac{1}{5}.$$
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