The question asks to find integer solutions to $4x^2 + 9y^2 = 72z^2$.
The solution states, as a first step, 'first notice that $x$ must be divisible by 3 and that $y$ is an even integer.'
Why is this the case? Does it have something to do with the factorization of 72 =($2^3 3^2$)?
All primitive solutions of $$ p^2 + q^2 = 2 r^2 $$ occur with all three odd, and $$ p = u^2 + 2 u v - v^2, \; \; \; q = -u^2 + 2 u v + v^2, \; \; \; r = u^2 + v^2 \; \; \; . $$ One of $u,v$ must be odd and the other even.
If you want positive variables you may just take absolute values. Your equation then has, with some integer $t$ that may be $1$ or larger, $$ x = 3pt, \; \; y = 2 qt, \; \; z = rt. $$