# All the natural numbers are of form $x^2+y^2+z^2+w^2+xy+zw$

Is there any natural number that cannot be written as $x^2+y^2+z^2+w^2+xy+zw$, where $x,y,z,w$ are integers? If there is none, how do I prove that all the natural numbers are of form $x^2+y^2+z^2+w^2+xy+zw$?

• For searching purposes, this would be called a universal quadratic form. Aug 4, 2014 at 0:50

$x^2 + xy+ y^2$ represents exactly the same numbers as $u^2 + 3 v^2.$ You are asking whether $$s^2 + 3 t^2 + u^2 + 3 v^2$$ is positive universal. It is.

More: a variation of quaternion multiplication shows that, when this form represents two numbers, call them $m,n,$ then it also represents $mn.$ Meanwhile, $$s^2 + 3 t^2 + u^2$$ is a regular ternary form, in particular it represents $1,2,3,$ and all numbers not divisible by $3,$ which includes all other primes. From the multiplication property, it follows that all (positive) numbers are represented.

For information about what numbers are represented by $s^2 + 3 t^2 + u^2,$ see the entry for $1,1,3$ in Dickson_Diagonal_1939.pdf at TERNARY. Acomplete proof is given as Theorem 89, pages 97-99, of Dickson's book.

• Every prime $\equiv 1\pmod 3$ is $u^2+3v^2$, so naturally I can represent multiplications of these primes via composition. How can I represent a $p\equiv -1\pmod 3$? Aug 4, 2014 at 1:36
• @IanMateus, $x^2 + y^2 + 3 z^2$ Aug 4, 2014 at 1:40

By the $15$ theorem, it suffices to check all naturals up to $15$. The margin is too small to fit my proof so the task is left to the reader.

• I think you would actually need to apply the 290 theorem here as the associated symmetric matrix contains halves. So your "too small proof" will need to be twice the size! (please correct me if I'm wrong here) Aug 4, 2014 at 1:06
• Haha, sorry, fixed the statement ;) Aug 4, 2014 at 1:13
• I should emphasize that this type of form, with odd crossed terms $xy$ and $zw,$ is not covered by the 15 Theorem; it is covered by the 290 Theorem, which is complete but may never be published. Aug 4, 2014 at 1:34
• @WillJagy: has anyone ever attempted a proof of the 15 and 290 theorem through the Jacobi triple product and Lambert series identities, as far as you know? Aug 4, 2014 at 1:47
• @JackD'Aurizio, not quite those words. series estimates are a key part of the 290 work; it is possible to say that a positive form in at least four variables does represent all numbers larger than a specific bound (different bound for each form), and then numbers up to that bound checked by computer. No need for the 15 Theorem, Bhargava's insight was that every example had to contain a regular ternary form as a sub-form. There is no shortcut for the 290 problem. Aug 4, 2014 at 2:04