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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?

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2 Answers2

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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).

user149792
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    Is finding that $3$ divides $x$, $y$, $z$ enough to show a contradiction? My thinking as follows. If this is the case, we can replace $x$ with $x' := x/3$, and similar for $y $ and $z$. We are left with an identical problem, so can again divide through by $3$, we can therefore do this infinitely which is of course impossible, hence contradiction. –  Sep 04 '15 at 21:43
  • @Numbers Showing $3,\vert,x$, $y$, and $z$ is a contradiction. The neater way to see this argument is that if we have a solution in $\mathbb{Q}$, we have a solution in $\mathbb{Z}$ (multiply through by denominator) and therefore can assume without loss of generality that $\text{gcd}(x, y, z) = 1$ (divide by any common factor). While this is not using the Hasse principle, it is showing that there is no solution in $\mathbb{Q}_3$ as well, hence if it makes you feel better, we can say we have shown there are no solutions in $\mathbb{Q}_3$ and hence by Hasse in $\mathbb{Q}$. – user149792 Sep 04 '15 at 21:47
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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.

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    the person asking does not seem to be aware of this, but the question is always about rational values for the variables. Indeed, since the thing is homogeneous, we might as well stick with integer variables. Here is a version with one variable put on the other side of the equals sign http://math.stackexchange.com/questions/27471/proof-of-legendres-theorem-on-the-ternary-quadratic-form – Will Jagy Sep 04 '15 at 21:37