Solve $x^2-y^2=2xyz$ 
Find all the integer solutions of
  $$x^2-y^2=2xyz.$$

I tried to factorise it, but I failed. Somebody help me.
 A: With $q=\dfrac xy$,
$$q^2-2zq-1=0.$$ 
The discriminant $z^2+1$ is never a perfect square so that $q$ is irrational except for the case $\color{green}{z=0}$ implying $\color{green}{x^2=y^2}$, to which we need to add the case of $y=0$, that is $\color{green}{x=y=0}$ ($z$ free).
A: Well you can always look at the equation as a polynomial of $x$:
$$x^2 - 2yzx - y^2=0$$
Now 
$$\Delta=4y^2z^2+4y^2=4y^2(z^2+1)$$
And finally
$$x_1=\frac{2yz-\sqrt{\Delta}}{2}=\frac{2yz-2y\sqrt{z^2+1}}{2}=yz-y\sqrt{z^2+1}$$
Analogously
$$x_2=yz+y\sqrt{z^2+1}$$
Now obviously we can pick $y$ arbitrarly and all we need is to either ensure that $\sqrt{z^2+1}$ is an integer or pick $y=0$. For the first case this can only be true for $z=0$ (I leave it as an exercise). For the second case we obtain solutions $x=y=0$ with $z$ arbitrary.
Thus only possible integer solutions are: ($x=\pm y$, $z=0$) and ($x=y=0$, $z$ arbitrary).
A: If $x=0$ or $y=0$ then the only solutions are $(0,0,b)$.
Otherwise, we may assume that $x$ and $y$ are relatively prime.
But then $x \mid y^2$ and $y \mid x^2$ imply $x=\pm 1$ and $y=\pm 1$ and so $z=0$.
Therefore, the only solutions are $(a,\pm a,0)$.
A: Well, $$x^2-y^2=2xyz $$ $$\Rightarrow x^2-2xyz-y^2=0$$ $$x = \frac {2yz \pm \sqrt {4y^2zz^2+4y^2}}{2} $$ $$\Rightarrow x = y (z \pm \sqrt {z^2+1}) $$
We can easily see that a trivial integer solution is $$(0,0,z) \,\, \forall z \in \mathbb Z $$ Since, for $x,y,z \in \mathbb Z $, the only integer $z$ for which $z^2+1$ will be a perfect square is zero, the other integer solutions are $$(a, \pm a,0) $$
Consider the case when $x,y,z \neq 0$. Then we have, $$x^2-y^2 =2xyz \Leftrightarrow \frac {x}{y}- \frac {y}{x} = 2z $$ Now when $x $ and $y $ are not equal as well as non-zero atleast one of the terms in the LHS is not an integer (why?) but the RHS is an integer (remember the condition?). So, this case yields no solution. 
Hope it helps. 
