When is the set of numbers represented by certain quaternary quadratic forms completely multiplicative? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T07:27:59Z http://mathoverflow.net/feeds/question/89054 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/89054/when-is-the-set-of-numbers-represented-by-certain-quaternary-quadratic-forms-comp When is the set of numbers represented by certain quaternary quadratic forms completely multiplicative? Will Jagy 2012-02-20T22:23:32Z 2012-02-21T05:29:31Z <p>Expired by this question <a href="http://mathoverflow.net/questions/88905/a-quadratic-form-represents-all-primes-except-for-the-primes-2-and-11" rel="nofollow">http://mathoverflow.net/questions/88905/a-quadratic-form-represents-all-primes-except-for-the-primes-2-and-11</a> 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.</p> <p>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 $21.$ </p> <p>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. </p> <p>So, that is the question, simple sufficient conditions for complete multiplicativity of integers represented by $$h(x,y,z,w) = f(x,y) + B g(z,w).$$</p> <p>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. </p> <p>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. </p> <p>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 <strong>principal genus</strong>? <strong><em>NO,</em></strong> not good enough, same results with $3 x^2 + 3x y + 4 y^2,$ of discriminant $-39,$ principal genus.</p> <p>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.</p>