If $(x_0, y_0, z_0, w_0)$ is the nontrival solution of $x^2-ay^2-bz^2+abw^2=0$, then $(x_0z_0-ay_0w_0, y_0z_0-x_0w_0, z_0^2-aw_0^2)$ is the nontrival solution of $x^2-ay^2-bz^2=0$.

Woah! What's the motivation?

Perhaps it is like cross product of corfficients.

How to get to the solution above: We first have to understand numbers of the form $x^2-ny^2$. The key idea is that the product of two such numbers is also such a number. In particular, we have the identity $\left(p^2-nq^2\right)\left(r^2-ns^2\right)=\left(pr-nqs\right)^2-n\left(ps-qr\right)^2$. With this in mind, the following solution seems slightly less magical. First rearrange to get $b\left(z^2-aw^2\right)=x^2-ay^2$. Now multiply both sides by $\left(z^2-aw^2\right)$. So we have $b\left(z^2-aw^2\right)^2=\left(x^2-ay^2\right)\left(z^2-aw^2\right)$. Now we use the identity above to get $b\left(z^2-aw^2\right)^2=\left(xz-ayw\right)^2-a\left(xw-yz\right)^2$. Simply bring all the terms to one side, and $\left(xz-ayw\right)^2-a\left(xw-yz\right)^2-b\left(z^2-aw^2\right)^2=0$. This is a solution of the form we wanted, so we are done. Note we could have factored out with $a$ instead of $b$, so we also get that $\left(xy-bwz\right)^2-a\left(y^2-bw^2\right)^2-b\left(xw-yz\right)^2=0$ is a solution. IMO2019 wrote: Perhaps it is like cross product of corfficients. This idea originates from the properties of determinants. For example, using determinants you can prove that the product of two numbers of the form $a^3+b^3+c^3-3abc$ is also of this form.

Thank you brilliant Kaskade.

https://artofproblemsolving.com/community/c3046h1046715_

A different solution: If $x^2 + abw^2 = ay^2 + bz^2$, then we have: $$a(xy+bzw)^2 + b(zx-ayx)^2 = (ay^2 + bz^2)^2.$$