I need help finding all positive integer solutions to the following Diophantine equation: $$\frac1x+\frac1y+\frac1z=\frac12$$
What I figured out so far was that we essentially need to find $3$ divisors, $a$, $b$, and $c$ of some number $k$ such that $$a+b+c=\frac k2$$This is so that when we divide $k$ on both sides, we get $$\frac ak+\frac bk+\frac ck=\frac12$$which is the same as $$\frac1{\frac ka}+\frac1{\frac kb}+\frac1{\frac kc}=\frac12$$We know that $k/a$, $k/b$, and $k/c$ are all integers because $a$, $b$, and $c$ are divisors of $k$, so by definition, they divide $k$ evenly. Furthermore, I found through experimentation that for any $k$, we can automatically assume that one of $a$, $b$, or $c$ is going to be $1$. This is because if all of them were $>1$, then the resulting fractions would simplify to form a sum that was already previously covered by a lower case.
So essentially, in order to find the solutions of this equation, we have to find two divisors $a$ and $b$ of some integer $k$ such that $a+b+1=k/2$. Then we would have the solution $$(x,y,z)=\left(\frac ka,\frac kb,k\right)$$
My question is: Is this a standard way of dealing with Diophantine equations of this form? Or is there a better way that I am unaware of? I would also like to know if this equation has a finite or infinite amount of solutions. From what I worked out, there intuitively seems to be infinite solutions, but I am not entirely sure if you can keep finding divisors that satisfy the conditions as $k$ grows bigger.
It was necessary to write the solution in a more General form:
$$\frac{t}{q}=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$$
$t,q$ - integers.
Decomposing on the factors as follows: $p^2-s^2=(p-s)(p+s)=2qL$
The solutions have the form:
$$x=\frac{p(p-s)}{tL-q}$$
$$y=\frac{p(p+s)}{tL-q}$$
$$z=L$$
Decomposing on the factors as follows: $p^2-s^2=(p-s)(p+s)=qL$
The solutions have the form:
$$x=\frac{2p(p-s)}{tL-q}$$
$$y=\frac{2p(p+s)}{tL-q}$$
$$z=L$$
It is enough to sort through the various options. $(tL-q)$
The overkill is insignificant. You can stop the search when the ratio becomes less than 2.