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Prove that $\{a^3+b^3+c^3-3abc\ |\ a,b,c \in \mathbb{Z}\}$ is closed under multiplication.

The easier version of this problem is to show that $\{a^2+b^2\ |\ a,b \in \mathbb{Z}\}$ is closed under multiplication, which can be done by $$\begin{align}(a^2+b^2)(c^2+d^2)&=(a+bi)(a-bi)(c+di)(c-di) \\ &=[(ac-bd)+(ad+bc)i][(ac-bd)-(ad+bc)i] \\ &=(ac-bd)^2+(ad+bc)^2. \end{align}$$

I think the proof would be similar to this, but I don't have any ideas except $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$ and then factoring the second factor with complex numbers, which seems too complicated.

Saturday
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    Hint: $a^3+b^3+c^3-3abc = (a+b+c)(a+\omega b + \omega^2 c)(a+\omega^2 b + \omega c)$, where $\omega = \dfrac{-1+\sqrt{3}i}{2}$ is the complex cubic root of $1$. – Zerox Jan 25 '23 at 13:15
  • https://www.linkedin.com/posts/safal-das-biswas-0a1b03255_find-some-time-to-read-this-activity-7003056330697555968-JE6d?utm_source=share&utm_medium=member_android –  Jan 25 '23 at 15:23
  • This paper suffices more general idea. I wrote this paper kinda 2 years back. –  Jan 25 '23 at 15:25
  • I had posted The question on aops kind in 2017, from Safal_db account on aops. I found the idea in 2017 only. –  Jan 25 '23 at 15:28
  • But however, couldn't do latex then, so write on words math type, but in between 2020-2021 I got proper generalization of the problem. –  Jan 25 '23 at 15:30
  • https://artofproblemsolving.com/community/u309941h1473224p8558914 –  Jan 25 '23 at 15:34
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    The reference you give deals with another issue : the fact that diophantine equation $a^3+b^3+c^3-3abc = n^2$ has (general ?) solutions $a=x^2-yz, \ b=y^2-zx, \ c=z^2 - xy, \ n=x^3+y^3+z^3-3xyz.$ – Jean Marie Jan 25 '23 at 17:12
  • One possible general solution yes. –  Jan 25 '23 at 21:50
  • I AM SORRY THERE A TYPO,, that should be $a=x^2+yz=a, b=z^2+xy, c=y^2+zx$ where $n=x^3+y^3+z^3-3xyz$ this is the correct one.Is one possible construction of infinitely many integer solution. MAY NOT GIVE EVERY POSSIBLE SOLUTION. –  Jan 25 '23 at 22:07

1 Answers1

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Let $$J=\pmatrix{0&1&0\\0&0&1\\1&0&0} \ \ \text{with} \ \ J^2=\pmatrix{0&0&1\\1&0&0\\0&1&0}\tag{1}$$

and

$$J^3=I \tag{2}$$

Consider the set $S$ of $3 \times 3$ "circulant" matrices :

$$M_{a,b,c}:=aI+bJ+cJ^2=\pmatrix{a&b&c\\c&a&b\\b&c&a}\tag{3}$$

Please note that $$\det(M_{a,b,c})=a^3+b^3+c^3-3abc\tag{4}$$

$S$ is stable by multiplication, due to property (2). Indeed :

$$(aI+bJ+cJ^2)(a'I+b'J+c'J^2)=$$

$$(aa'+bc'+b'c)I+(ab'+ba'+cc')J+(ac'+bb'+a'c)J^2\tag{5}$$

In other words:

$$M_{a,b,c} \times M_{a',b',c'}=M_{(aa'+bc'+b'c),(ab'+ba'+cc'),(ac'+bb'+a'c)}\tag{6}$$

Taking the determinant on both sides of (6) and applying (4), one can conclude to the stability by multiplication. Of course, the above operations, when done with entries in $\mathbb{Z}$, give back entries in $\mathbb{Z}$.

Remark 1: Properties (1) and (2) for $J$ have a natural interpretation as (cyclic) permutation matrices.

Remark 2: One can check easily that set $S$ has all the required properties for being a (unitary) ring for matrix addition and multiplication. Can this ring be turned into a field ? If this was possible, by isomorphism, (the additive group of $S$ is isomorphic to the group $(\mathbb{R}^3,+)$), it would imply that $(\mathbb{R}^3,+,\times)$ can be turned into a field. But we know it is impossible by a theorem due to Frobenius. What is missing ? The fact that some elements haven't an inverse, more precisely those with

$$\underbrace{x^3+y^3+z^3-3xyz}_{\det M_{x,y,z}}=0\tag{7}$$

A classical factorization of the LHS is

$$(x+y+z)((x-y)^2+(y-z)^2+(z-x)^2)=0\tag{8}$$

A consequence of (8) is that the set of non-invertible matrices $M_{x,y,z}$ is the union of

$$\begin{cases}\text{a plane : } x+y+z=0\\ \text{a line : } x=y=z\end{cases},\tag{9}$$

otherwise said to 2 families of matrices :

$$\begin{cases}\pmatrix{x&y&-x-y\\-x-y&x&y\\y&-x-y&x} \text{rank 2 matrices,}\\ \pmatrix{x&x&x\\x&x&x\\x&x&x} \text{rank 1 matrices} \end{cases}\tag{9'}$$

I have investigated in a graphical way the relative "positions" of the families of matrices $M_{x,y,z}$ such that

$$x^3+y^3+z^3-3xyz = k$$

for different values of constant $k$.

The result is displayed on the figure below under the form of "russian dolls" surfaces, a kind of "foliation" of space $\mathbb{R^3}$ ; in a progressive manner, these surfaces "take their distance" wrt the degenerate case, with plane $x+y+z=0$ as their asymptotic plane, and a chimney-like part around the line $x=y=z$ (featured in black).

enter image description here

Fig. 1 : Surfaces $x^3+y^3+z^3-3xyz = k$ for $k=-8,-4,0,4,8$.

Edit : All these surfaces are surfaces of revolution around the line defined in (9) as established in this answer.

Jean Marie
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