Infinitely many integer triples $(x, y, z)$ satisfying $x^2 + 2y^2 = 3z^2$

Question

My book mentions a proof involving a 1:1 correspondence with rational slopes through the point $(1, 1)$, I guess for every value of $z$. I'm interested in this approach but I don't know how to make it work.

Otherwise, how would I go about proving this?

Answer 1

We may find all the rational points on $X^2+2Y^2=3$ by just considering that $(1,1)$ is one of them. Let us consider a line through $(1,1)$ with rational slope $m$ and its second intersection with the given ellipse: by Vieta's theorem, it is a rational point. Namely it is

$$ (X,Y) = \left(\frac{2m^2-4m-1}{2m^2+1},-\frac{2m^2+2m-1}{2m^2+1}\right) $$ and this gives a parametrization of all the rational points on our ellipse, with the only exception of $(X,Y)=(1,-1)$. By setting $m=\frac{p}{q}$ we may easily derive the identity

$$(2p^2-4pq-q^2)^2+ 2(2p^2+2pq-q^2) = 3(2p^2+q^2)^2 $$ proving the claim about the existence of infinite primitive solutions.

Answer 2

If $(x_0,y_0,z_0)$ is a solution to this problem, then is also $x=x_ot$, $y=y_ot$ and $z=z_ot$ for each integer $t$. So you have to find this one solution $(x_0,y_0,z_0)$ and this is not difficult to do, since $(1,1,1)$ will do.