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can anyone prove that every natural odd number can be represented as the form $a^2+b^2+2c^2$ where $a,b$ and $c$ are nonnegative integers?

I've thinked on this problem for a long time, but I couldn't solve it. I'd really appreciate your help.

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The three-squares theorem I use in my answer is proved in detail in Dickson's 1939 book. It is also proved in John Horton Conway, The Sensual Quadratic Form, (1997), using the clever Aubry-Davenport-Cassels trick, but first uses strong Hasse-Minkowski. So Dickson's treatment is likely to seem easier, if you can accept Dirichlet's celebrated theorem on primes in arithmetic progressions. Also Conway style in Jean-Pierre Serre, A Course in Arithmetic (1973 in English) – Will Jagy Sep 29 '11 at 19:35
up vote 3 down vote accepted

An actual proof is in a 1939 book (Theorem 86, page 96) by Leonard Eugene Dickson called Modern Elementary Theory of Numbers. I give Dickson's list (pages 111-113) of "diagonal" regular ternary forms at DIAGONAL. Bhargava's work depends heavily on existing results such as these. Alright, I put Manjul's article, with a preface by Conway, at BHARGAVA. Finally, for a positive quadratic form to represent all numbers, it must have at least four variables. However, it is possible to represent all odd numbers with only three variables. Kaplansky identified all possible such ternary forms in KAPLANSKY. Kap gave 23 ternary forms that seemed to work, 19 he could prove (or had already been proved, such as your $x^2 + y^2 + 2 z^2$) along with four plausible candidates. I proved one of the four, but three are still uncertain. Henri Cohen likes to start his classes by mentioning one of those three forms, pointing out that we cannot prove what numbers it represents.

So, a few cautions. The 15 Theorem and the 33 Theorem are about "classically integral" forms, all mixed terms $x_i x_j$ have even coefficients. There is a 290 theorem by Bhargava and Hanke about positive forms that represent all numbers. In this case, analytic methods are able to finish the problem. It is not clear the 290 result will ever be published, it has been submitted and withdrawn once already.

Finally, there is no theorem giving a bound that guarantees a positive form does indeed represent all odd positive integers, as soon as we allow forms that are not classically integral. There cannot be such a theorem as long as there is no proof that Kaplansky's three stubborn forms really work.

Plenty of other information is at TERNARY_SITE

EDIT: if one is willing to take as an axiom the three-squares theorem, Legendre 1798, Gauss 1801, it is an easy additional step. So, Legendre, a positive integer $r$ can represented as $r = x^2 + y^2 + z^2$ if and only if $r$ is not of type $4^k (8m+7),$ with integers $k,m \geq 0.$ So, take an odd positive number $n.$ We can see that $2n$ satisfies the three-square theorem, and we have $$ 2 n = x^2 + y^2 + z^2.$$ The three numbers cannot all be odd, so, perhaps by renaming variables, demand that $z$ be even, or $z = 2 c.$ So far we have $$ 2 n = x^2 + y^2 + 4 c^2.$$ Next, $x,y$ are either both odd or both even, anyway both $x+y$ and $x-y$ are both even. Take $$ a = \frac{x-y}{2}, \; \; b = \frac{x+y}{2} $$ and we get $$ n = a^2 + b^2 + 2 c^2 $$ in integers.

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nice proof using three-squares theorem. thanks :) – Goodarz Mehr Sep 29 '11 at 11:41

Bhargava has solved the problem of which integral quadratic forms represent all odd naturals: it's sufficient that the form represent 1, 3, 5, 7, 11, 15, and 33. It's easy to check that your form represents these numbers, and hence all odd natural numbers.

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isn't there an elementry olympiad-level proof for the question? – Goodarz Mehr Sep 28 '11 at 16:27
@goodarz: I doubt it. This result easily implies Lagrange's theorem that every positive integer is a sum of four squares, which I believe is well above the olympiad level. – Chris Eagle Sep 28 '11 at 17:00
Trivial remark: you mean Manjul Bhargava (not Barghava). – David Loeffler Sep 28 '11 at 17:19
@David: bleh, thanks – Chris Eagle Sep 28 '11 at 17:22

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