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Expired by this question A quadratic form represents all primes except for the primes 2 and 11. I would like to know some simple sufficient conditions for when the set of numbers integrally represented by a positive four-variable quadratic form is completely multiplicative, with $$ h(x,y,z,w) = f(x,y) + B g(z,w), $$ while $f,g$ are positive primitive binary quadratic forms of the same discriminant, and integer $B \geq 1.$ The only example I am completely sure about is $$ h(x,y,z,w) = x^2 + A y^2 + B z^2 + A B w^2. $$ The reason this one is easy is that we can construct quaternions $$ x + y i \sqrt A + z j \sqrt B + w k \sqrt {A B}, $$ so the quadratic form is the norm and multiplication is built in. In particular, taking $B=1,$ we have the principal form repeated.

Noam Elkies pointed out that the condition of complete multiplicativity does not hold for repeated binary $3 x^2 + 7 y^2,$ as $$ h(x,y,z,w) = 3 x^2 + 7 y^2 + 3 z^2 + 7 w^2$$ represents $3$ and $7$ but not $21.$

I had this feeling that the fact the $3 x^2 + 2 x y + 4 y^2$ was in a discriminant of one genus and class number three would suffice for the form in question 88905. By itself, the binary form represents a completely multiplicative set, as its cube in the class group is the identity but its square is simply its own opposite class. But I am not entirely sure about the quaternary in question 88905. Furthermore, I am getting other examples, class number five and so on.

So, that is the question, simple sufficient conditions for complete multiplicativity of integers represented by $$ h(x,y,z,w) = f(x,y) + B g(z,w).$$

EDIT, 2:56 pm: Note that the numbers represented by this $f(x,y) + B g(z,w)$ can always be multiplied by any number represented by the principal form of the same discriminant, as this holds true separately for $f,g.$ For this reason, if the form in question 88905 is completely muliplicative (as seems very likely) it suffices to consider primes $p \neq 2,11$ along with $2p$ for $p \neq 2,11.$ Noam has already done such the first case.

EDIT TOOOO, 5:06 pm: Noam says that, in question 88905, his answer extends to show that the set of numbers represented is indeed multiplicative, but he does not expect there to be a direct proof of that. So one might say that I am asking for situations where there is a direct proof. For example, there are plenty of forms in four variables that represent all positive integers (if they are integer-matrix, they just need to represent the numbers up to 15, if not integer matrix then we need them to represent up to 290 to be sure). Anyway, a positive form that represents all numbers integrally is "completely multiplicative" in the way I define it, but that does not mean there is any nicer proof.

EDIT THREE, 9:18 pm: I have been running examples. Another counterexample is repeating the binary $3 x^2 + x y + 3 y^2,$ of discriminant $-35,$ the quaternary represents $3,7$ but not $21.$ However, the class number of the principal genus does not seem to matter. So a reasonable revised question is, how about when both $f,g$ are in the principal genus? NO, not good enough, same results with $3 x^2 + 3x y + 4 y^2,$ of discriminant $-39,$ principal genus.

I have no idea what is going on. Many of these clearly work, some just don't. If they fail, it seems to be with very small numbers.

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Are you expired or inspired? (This is a real question, designed to improve my understanding of English vocabulary as well as your current condition. Sorry for the off-topicness.) – Jonathan Chiche Feb 21 '12 at 0:28
@Jonathan, inspired is the usual term. And expired usually means dead, with the single exception of actual breathing, inspiration and expiration. I see many questions that begin "Inspired by" and I think the phrase is, at best, overused. I see, I looked it up, medical people use those terms for breathing, most people say inhalation and exhalation. See – Will Jagy Feb 21 '12 at 0:28
^^ Heh. Typical Will. – Koundinya Vajjha Feb 21 '12 at 1:18
Dear Will: Thanks for the explanation. Actually, there are similar pairs of words in French, and "inspiré" is slightly overused too, in the same fashion (as far as as I can tell) as in English. However, I am unsure as to how people would understand a sentence such as "Je suis expiré". I will have to try it some day. – Jonathan Chiche Feb 21 '12 at 1:24
An obvious sufficient condition is that the extension $h_{\mathbb{Q}}$ is multiplicative (in other words it is isotropic or similar to Pfister form) and h satisfies the hypotheses of Cassels-Davenport theorem. – Yazdegerd III 0 secs ago – Name May 4 '13 at 16:05

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