Describe all integral solutions of the equation $x^2 + y^2 = 2z^2$ such that $x,y,z > 0$, gcd$(x,y,z) = 1$, and $x > y$. As the title states, the question tasks me with finding all the integral solutions of the equation under the specified constraints. I have an idea of where to start due to a somewhat similar problem in my notes, but I'm having trouble adapting it to this new equation, and knowing where I need to change my process. My professor has provided the hint: "Use the circle $^2 + ^2 = 2$ and the lines
passing through the point $(1,1)$." Here's what I have so far:
$x^2 + y^2 = 2z^2$
$\frac{x^2}{z^2} + \frac{y^2}{z^2} = 2$
Let $X = \frac{x}{z}$ and $Y = \frac{y}{z}$
$X^2 + Y^2 = 2$
So we now have a circle with an origin at $(0,0)$ and radius of $\sqrt{2}$. I then drew up the circle and the line passing through $(1,1)$.

So the slope $\lambda$ of this line would be: $\lambda = \frac{Y-1}{X-1}$. This is the part where I get lost, in class we went off on a tangent related to these problems and I'm having trouble knowing exactly how to proceed. I have an idea of what the final answer will look like. The form we found for $x^2 + y^2 = z^2$ was: $(x,y,z) = (a^2 - b^2, 2ab, a^2 + b^2)$ Any help would be greatly appreciated!

Since you have worked out the slope, work out the equation of the $x$-coordinates of the two intersections of the line and the circle, then since you know the coord of one, use Viete's theorem.
