Quadratic form has non-trivial zero?

Question

For each of the following quadratic forms, determine whether the form has a non-trivial zero (we do not need to exhibit it):

1. $f(x, y, z) = 2x^2 + 3y^2 - 6z^2$;

2. $g(x, y, z) = 2x^2 + 3y^2 - 10z^2$;

3. $h(x, y, z) = x^2 + y^2 - 64z^2$.

I'm confused on this problem and not very sure how to start. Could anyone give me a tiny hint?

Answer 1

For the first quadratic equation choose an arbitrary $x$ and $y$ and then take

$$z=\sqrt{\frac1{6}(2x^2+3y^2)}$$ you get $(x,y,z)$ a non trivial solution.

Answer 2

Here are two useful facts:

1. Hidden in Hasse's lemma on conics (I am not sure if this is the standard name) is the fact that there are always solutions in $\mathbb{Q}_p$ if $p$ does not divide any coefficients. (Hence, we just need to check a few primes.)
2. There can be no solutions for an even number of primes (where we are counting $\infty$ as a prime, $\mathbb{Q}_\infty = \mathbb{R}$). So we just need to check if there are no nontrivial solutions for all but one prime.

With this in mind, here is a hint for the first quadratic form $f(x, y, z)$. Try to find a contradiction to a solution in $\mathbb{Q}_3$ by looking at $\text{mod }3$, see what you can deduce, and use that by looking at $\text{mod }9$ (and hopefully at this point you should find your argument why no nontrivial solution exists).