Extension of Dickson's theorem on integers of the form $a^2+b^2+2c^2$

Question

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 $A \in M_3(\mathbb{N})$.

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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.

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Computation

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

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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

Answer 1

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 $F$.

Answer 2

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