All integer solutions to diophantine equation: $x^2+p y^2=z^2$? I would like to find all integer solutions to the Diophantine equation
$$
x^2+p y^2=z^2
$$
where $p\ge2$ is a given prime number. Also prove that my (probably parametric similar to Pythagorean triples $x=a^2-b^2,\ y=2ab,\ z=a^2+b^2$) solution gives all of the.
I tried to move $x^2$ to the right hand side and write $py^2=(z-x)(z+x)$. then assume two cases: (1)  $\ \ p|z-x$ , (2)  $ \ p|z+x$. then I wrote some conditional relation depending on $y$. but I am not satisfied with that. Any help would be appreciated?
 A: This is no different from the Pythagorean case.  To classify integer solutions, look instead for rational solutions of $$x^2+py^2=1$$
Thus we want rational points on an ellipse.  One solution is $P =(-1,0)$.  Stereographically project from that point (that is, given a point $(0,h)$ we extend the line connecting it to P and find its intersection with the ellipse).  Elementatry algebra shows that that the point on the ellipse we find this way is rational if and only if h is rational.  Explicitly, we get the solution $$\left(\frac{1-ph^2}{1+ph^2},\frac{2h}{1+ph^2}\right)$$
Clearing denominators and letting $h=\frac ab$ we get the general solution $$\left(b^2-pa^2,2ab,b^2+pa^2\right)$$
By construction, this gives all rational solutions up to an integer multiplier.  As a commenter( WillJagy) correctly remarked, this method can generate solutions with common factors which must then be canceled to obtain primitive solutions.  
A: $$py^2=z^2-x^2=(z+x)(z-x)$$
$p$ is an odd prime, then it divides one of the factors.
Then,WLOG:
$$\dfrac{z+x}{p}=g^{2-t}a^2$$
$$z-x=g^tb^2$$
This yields
$$\pm z=\dfrac{pg^{2-t}a^2+g^tb^2}{2}$$
$$\pm x=\dfrac{pg^{2-t}a^2-g^tb^2}{2}$$
$$\pm y=gab$$
Where $a,b,g$ are all integers.
