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I was reading about the Hilbert residue symbol, and the discussion of it starts out with the assumption that we can reformat any ternary quadratic form over the integers into the form

$ax^2+by^2-z^2, a,b\in\mathbb{Z}$

by first diagonalizing and then rescaling. I managed to work out the diagonalization part, but am lost on the rescaling. I understand how to make sure that the 3 coefficients are coprime and squarefree, but after that, for example when left with something like

$3x^2+5y^2-7z^2$

I am lost.

jana_d
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  • You are interested, for the Hilbert symbol, not in a quadratic from $Q$ per se, but in solvability of the equation $Q = 0$. And for that you can multiply by any nonzero scalar without changing solvability. For your example, $Q = 0$ is the same as $7Q = 0$, and $7Q = 21x^2 + 35y^2 -(7z)^2$, so solvability of your quadratic form equal to $0$ in rational numbers that are not all zero is the same as solvability of $21x^2 + 35y^2 - z^2 = 0$. – KCd Jul 05 '14 at 16:16

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For a full treatment of a ternary using the Hilbert Norm Residue symbol, see my two answers at Isotropy over $p$-adic numbers

Got to admit, I never thought of the simple manipulation in the comment by KCd; quite possible that is all you really need. It is exactly what Cassels is saying on page 59, proof of Lemma 2.5, I did not put it together.

Meanwhile, many books, including Cassels, give full detail on the theorem of Legendre saying when a diagonal ternary $a x^2 + b y^2 + c z^2$ with integer $a,b,c$ has a nontrivial integer zero $(x,y,z).$ Page 80 in Cassels, Theorem 4.1, where he comments (middle page 82) that conditions (ii), (iii) are unnecessary if replaced with the hypothesis that the form be indefinite. I like the wording in condition (i), he makes it quite clear that each odd prime dividing any of $a,b,c$ matters; he is already demanding the product $abc$ squarefree, so at most one is even. Here is an MSE question with a different wording of Legendre, the insistence on relatively prime coefficients is really the whole story in this version: Proof of Legendre's theorem on the ternary quadratic form

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