11
$\begingroup$

Theorems V in this paper of L.E. Dickson states that the following two sets are equal. $$E=\{a^2+b^2+2c^2 \ | \ a,b,c \in \mathbb{Z}\} \ \text{ and } \ F=\mathbb{N} \setminus \{4^k(16n+14) \ | \ k,n \in \mathbb{N}\}.$$ Let $\mathbb{A}$ denote $\mathbb{N}$ or $\mathbb{Z}$. Consider the following set: $$E(\mathbb{A}) = \left\{\frac{1}{2}\|u+v \|^2 \text{ with } u,v \in \mathbb{A}^3 \text{ and } \|u \| = \|v \| \right\}.$$

Let $u= (a,b,c)$, $v = (b,-a,c) \in \mathbb{Z}^3$. Then $\|u \| = \|v \|$ and $\frac{1}{2}\|u+v \|^2 = a^2+b^2+2c^2.$
It follows that $F= E \subseteq E(\mathbb{Z})$. Now, by Legendre's three-square theorem, $E(\mathbb{Z}) \subset F$ also.
Then, we have an extension of Dickson's theorem as $E(\mathbb{Z}) = F$. Now, what about $E(\mathbb{N})$?

Take $u=v \in \mathbb{N}^3$, then $\frac{1}{2}\|u+v \|^2 = 2 \|u \|^2$, so by Legendre's three-square theorem, $E(\mathbb{N})$ contains the even part $F$. The computation below shows that $E(\mathbb{N})$ contains every odd number less than $95362$, except those in $I=\{ 5, 23, 29, 65, 167 \}$, suggesting that $E(\mathbb{N}) = F \setminus I$.

Question: Is it true that, for $u,v \in \mathbb{N}^3$ with $\|u \| = \|v \|$, the form $\frac{1}{2} \|u+v \|^2$ covers every odd number, except those in $\{ 5, 23, 29, 65, 167 \}$?

Application: this answer proves that the form $\| A\|^2$ covers every natural number for $A \in M_3(\mathbb{Z})$.
A positive answer to the above question would prove this result for $A \in M_3(\mathbb{N})$.


For the convenience of the reader, the answer of Philipp Lampe (of what was Question 1 in a previous version) was incorporated in the post.


Computation

sage: L=cover(135)
sage: set([2*i+1 for i in range(47681)])-set(L)
{5, 23, 29, 65, 167}

Code

# %attach SAGE/3by3.spyx

from sage.all import *

cpdef cover(int r):
    cdef int a1,a2,a3,b1,b2,b3,x,n
    cdef list L
    L=[]
    for a1 in range(r):
        for a2 in range(a1+1):
            for a3 in range(a2+1):
                x=a1**2+a2**2+a3**2
                for b1 in range(isqrt(x)+1):
                    for b2 in range(isqrt(x-b1**2)+1):
                        for b3 in range(isqrt(x-b1**2-b2**2)+1):
                            if a1**2+a2**2+a3**2==b1**2+b2**2+b3**2:
                                n=((a1+b1)**2+(a2+b2)**2+(a3+b3)**2)/2
                                if is_odd(n) and not n in L:
                                    L.append(n)
    return L
$\endgroup$
  • $\begingroup$ In this post, $\mathbb{N}$ is the set of non-negative integers. $\endgroup$ – Sebastien Palcoux Nov 3 '18 at 14:59
16
$\begingroup$

Answer to Question 1. Yes, $E(\mathbb{Z})=F$.

The inclusion $F\subseteq E(\mathbb{Z})$ follows from $E\subseteq E(\mathbb{Z})$ and Dickson's result $E=F$. It remains to show $E(\mathbb{Z})\subseteq F$. Pick $n\in E(\mathbb{Z})$. By definition there exist $u,v\in\mathbb{Z}^3$ such that $\lVert u\rVert=\lVert v\rVert$ and $n=\lVert u\rVert^2+\lvert u\cdot v\rvert$. Then $$2n =\lVert u\rVert^2+ \lVert v \rVert^2+ 2\lvert u\cdot v\rvert = \lVert u+v\rVert^2 $$ is a sum of three squares. Legendre's three-square theorem implies that $2n$ cannot be written as $4^a\left(8b+7\right)$ with $a,b\geq 0$. From this we can conclude that $n$ must belong to $F$.

$\endgroup$
9
$\begingroup$

given your interest: the list of all $A x^2 + B y^2 + C z^2$ with ordered positive coefficients, such that the represented numbers can be described by congruences

Ummm. Allowing mixed terms, all 913 (probably) regular positive forms

enter image description here

$\endgroup$
  • $\begingroup$ Then, what is, for example, $ \{ x^2+y^2+z^2+xy+xz+yz \ | \ x,y,z \in \mathbb{N} \} $? $\endgroup$ – Sebastien Palcoux Oct 28 '18 at 14:26

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.