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What is a Quaternary Quadratic Form? I've looked for a definition online and cannot find a precise clear definition.

I am not taking a course. Just reading about quadratic forms.

Thank you.

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    $\begingroup$ A quaternary quadratic form is a quadratic form in 4 variables. (A binary form is one in 2 variables, a ternary in 3, etc.) $\endgroup$ – anomaly Jun 28 '14 at 15:55
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    $\begingroup$ This sort of terminology is unnecessarily confusing. I already think ternary quadratic form is confusing. Just say "quadratic form in $n$ variables" where $n$ is whatever. $\endgroup$ – Qiaochu Yuan Jun 28 '14 at 18:09
  • $\begingroup$ @anomaly: if you write your comment as an answer you can get your reward for having the best answer. $\endgroup$ – ReverseFlow Dec 13 '14 at 22:43
  • $\begingroup$ @jvargas: It's pretty short for a full answer, but sure, I won't turn down reputation points. :) $\endgroup$ – anomaly Dec 14 '14 at 0:09
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A quaternary quadratic form is a quadratic form in 4 variables. (A binary form is one in 2 variables, a ternary in 3, etc.)

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For positive ones see

http://www.math.rwth-aachen.de/~Gabriele.Nebe/LATTICES/nipp.html

http://www.math.rwth-aachen.de/~Gabriele.Nebe/LATTICES/nipp.html#form

http://www.math.rwth-aachen.de/~Gabriele.Nebe/LATTICES/d4to457.html

I do not know of any tables of indefinite quaternaries. The interesting examples are of this type: $$ w^2 - 2 x^2 + 3 y^2 - 6 z^2. $$ For real irrational values of the variables, it is easy to make this add up to $0,$ just take $w=\sqrt 2, x=1, y=0,z=0.$ For integers (and so rationals) the sum cannot be $0$ unless all four of $w,x,y,z=0.$ This is simply because, for integers, $w^2 - 2 x^2 + 3 y^2 - 6 z^2 \equiv 0 \pmod 9$ means $w,x,y,z \equiv 0 \pmod 3. $ If there were any nontrivial integer expression $w^2 - 2 x^2 + 3 y^2 - 6 z^2=0,$ there would also be one with $\gcd(w,x,y,z) = 1;$ but we see that this is not possible.

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  • $\begingroup$ Alternatively: Suppose $w, x, y, z$ is a solution with minimal $|w|$. Modulo 3, the $y$ & $z$ terms vanish. $w^2=0$ or 1; $2x^2=0$ or 2; they are equal so (as you say) $w=x=0$. Now substitute $y, z, w/3, x/3$ for $w, x, y, z$. Repeat the argument. This gives a smaller solution $w/3, x/3, y/3, z/3$ contradicting minimality. $\endgroup$ – Rosie F Feb 13 at 21:57

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