You can do it by intersecting lines with the implied circle $x^2+y^2=2$. Or you can see that $a=\frac{x+y}{2} , b=\frac{x-y}{2}$ fulfill $a^2+b^2=z^2$, so it is reduced to pythagorean tripels ($a,b$ will be integral because both, $x,y,$, will be odd).

alexilic wrote: Solve the following equation in integers with gcd (x, y) = 1 $x^2 + y^2 = 2 z^2$ Actually, the problem says gcd (x, y, z)=1. http://www.oma.org.ar/enunciados/con5.htm

Thats totally the same

wouldnt it be easier to identify integer points on the curve(parabola) $2z^2$ and after that set the radius of the circle to be the distance from origin to that integer coordinates point on the parabola ?

ZetaX wrote: Thats totally the same can you explain me why?

José wrote: ZetaX wrote: Thats totally the same can you explain me why? Well, if gcd(x,y) = 1 then gcd(x,y,z) = 1, and if some prime divides both x and y, putting that into the equation gives that it divides z as well..

TripleM it divides z as well or the prime is 2

spx2 wrote: TripleM it divides z as well or the prime is 2 It works for p=2 as well (4 divides the LHS, so 2 divides z^2, so 2 divides z).

$ x=(a^2-2ab-b^2)c,y=(b^2-2ab-a^2)c,z=(a^2+b^2)c, (a,b)=1$ if a and b odd, then c=1/2, else c=1.

You are wrong because $ (x,y,z)$ parewise relative prime. You can use complex integer or geometric method :Find the rational point on the curve $ x^2+y^2=2$

TTsphn wrote: Find the rational point on the curve $ x^2 + y^2 = 2$ How can we do it?Show your complete solution.Please explain in details.Thank you.

Yes we has the geometric method. If Because $ (1,1)$ is an integer point on the curve by viete law we can prove that : 1)$ d: a(x - 1) + b(y - 1) = 0 (a,b\in Q)$ is a straight throw across $ (1,1)$ and cut the curve at two point $ (x_0,y_0)$ then $ (x_0,y_0)$ is a rational ponit. 2)The if the straight go to across $ (x_0,y_o)$ is a rational on curve then $ a,b\in Q$ So we must solve two equation $ a(x - 1) + b(y - 1) = 0$ $ (x^2 + y^2 = 2$ where $ (a,b)\in Q$ It is quite easy .