Number of solutions of $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=1$ Find the number of solutions $(a,b,c,d)$ such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=1$,$\enspace$$a,b,c,d\in \mathbb{N}$ and $a<b<c<d$.

I have got a solution:
$$\frac{1}{2}\cdot(\frac{1}{2}+\frac{1}{3}+\frac{1}{6})+\frac{1}{2}=\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{12}=1$$
But how to get other solutions.
I received total $6$ solutions from Wolfram Alpha.


*

*Show $a\le4$.

*Run through various possible values of $a$ and $b$.

*To solve ${1\over c}+{1\over d}={1\over4}$ for example,
$${1\over c}+{1\over d}={1\over4}\\
4d+4c=cd\\
(c-4)(d-4)=16$$
Now just run through the factors of 16.

