Solving the diophantine equation $1/x+1/y+1/z=1/2$ 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.
 A: A common first step in solving such a problem is to clear denominators to get a polynomial equation. Multiplying by $2xyz$ yields
$$2yz+2xz+2xy=xyz,\tag{1}$$
so every solution to $(1)$ with $xyz\neq0$ yields a solution to the original equation, and conversely every solution to the original equation yields a solution to $(1)$. Note that all solutions with $xyz=0$ are of the form
$$(x,0,0),\qquad (0,y,0),\quad\text{ or }\quad (0,0,z).$$
A common next step is to find some divisibility restrictions on $x$, $y$ and $z$, and to take out common factors. This is possible because we have reduced the problem to finding the zeros of a polynomial.
Let $(x,y,z)$ be a solution with $xyz\neq0$. Let $D:=\gcd(x,y,z)$ so that $x=DX$, $y=DY$ and $z=DY$ for integers $X$, $Y$ and $Z$ with $\gcd(X,Y,Z)=1$. Then
$$2YZ+2XZ+2XY=DXYZ.\tag{2}$$
From $(2)$ we immediately see that $X$ divides $2YZ$, that $Y$ divides $2XZ$ and that $Z$ divides $2XY$. This is quite the restriction on $X$, $Y$ and $Z$, which becomes more clear by considering
$$U:=\gcd(Y,Z),\qquad V:=\gcd(X,Z),\qquad W:=\gcd(X,Y).$$
First note that $U$, $V$ and $W$ are pairwise coprime because $\gcd(X,Y,Z)=1$.
Then we have
$$X=VWX',\qquad Y=UWY',\qquad Z=UVZ',$$
for integers $X'$, $Y'$ and $Z'$ that are also pairwise coprime. Now we see why the divisibility relations on $X$, $Y$ and $Z$ that we found are so restrictive: We saw that $X$ divides $2YZ$, where now
$$X=VWX'\qquad\text{ and }\qquad 2YZ=2U^2VWY'Z',$$
and $X'$ is coprime to $U$, $Y'$ and $Z'$. It follows that $X'$ divides $2$, and entirely analogously that $Y'$ and $Z'$ divide $2$. In fact, because $X'$, $Y'$ and $Z'$ are pairwise coprime, we see that $X'Y'Z'$ divides $2$.
From here we can reconstruct all solutions to $(1)$. We have just shown that $x$, $y$ and $z$ must be of the form
\begin{eqnarray}
x&=&DX&=&DVWX',\\
y&=&DY&=&DUWY'\\
z&=&DZ&=&DUVZ',
\end{eqnarray}
where $U$, $V$ and $W$ are pairwise coprime positive integers, and $X'Y'Z'$ divides $2$, and $D$ is a positive integer. Plugging this into $(1)$ yields
$$2D^2U^2VWY'Z'+2D^2UV^2WX'Z'+2D^2UVW^2X'Y'=D^3UVWX'Y'Z'.$$
Because $X'Y'Z'$ divides $2$ we can divide out $D^2UVWX'Y'Z'$ to get
$$D=\frac{2U}{X'}+\frac{2V}{Y'}+\frac{2W}{Z'}.$$
This means we can choose any three pairwise coprime positive integers $U$, $V$ and $W$, and three integers $X'$, $Y'$ and $Z'$ such that $X'Y'Z'$ divides $2$, and set
$$D:=\frac{2U}{X'}+\frac{2V}{Y'}+\frac{2W}{Z'},$$
which is then also an integer, to get a solution
$$(x,y,z)=(DVWX',DUWY',DUVZ').$$
Moreover, all solutions are of this form.
A: 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.
