Solving Diophantine equation $1/x^2+1/y^2=1/z^2$ How can we find positive integers solutions $(x,y,z)$, where $\gcd(x,y,z)=1$ for the equation:
$$1/x^2+1/y^2=1/z^2$$
Can we conclude that $x$ and $y$ are not coprimes for it to have solution?
 A: Multiply both sides by $x^2y^2z^2$.
Then you get
$$y^2z^2+x^2z^2=x^2y^2$$
Now use that each of $x^2, y^2,z^2$ divide two of the terms hence the third.
Added Here is the rest of the solution. Let $a=gcd(x,y), b=gcd(x,z), c=gcd(y,z)$.
Then, $gcd(a,b)=1$ and hence $ab|x$. We claim $ab=x$.
Indeed write $x=abd$. Assume by contradiction that $d \neq 1$ and let $p|d$, $p$ prime.
As $x | yz$ we have $abd | yz \Rightarrow d | \frac{y}{a}\frac{z}{b}$.
Then $p$ divides either $\frac{y}{a}$ or $\frac{z}{b}$.
But then, in the first case $pa |x,y$ while in the second $pb | x,z$ contradicting the $gcd$.
Therefore $x=ab$. The  same way you can prove that $y=ac, z=bc$.
Replacing in the above equation you get
$$a^2b^2c^4+a^2b^4c^2=a^4b^2c^2$$
or
$$c^2+b^2=a^2$$
this shows that $(c,b,a)$ is a primitive Pytagoreal triple and 
$$x=ab \\
y=ac \\
z=bc$$
A: Although the question's old, I'll leave a solution which I believe is a correct for someone who comes along.
Obviously, $x \ge y > z$.
You should see why $\gcd(x,y)= 1$. 
So, now:
$$\frac{1}{y^2} = \frac{x^2 - z^2}{x^2z^2}$$
Hence, $y^2|x^2z^2 \implies y|xz \implies y|z \text{ since } x \text{ and } y \text{ are relatively prime.}$
$y \le z$.
Contradiction.
