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 (where $u \cdot v$ denotes the usual dot product): $$E(\mathbb{A}) = \{\|u \|^2 + |u \cdot v| \text{ such that } u,v \in \mathbb{A}^3 \text{ and } \|u \| = \|v \|\}.$$$$E(\mathbb{A}) = \left\{\frac{1}{2}\|u+v \|^2 \text{ with } u,v \in \mathbb{A}^3 \text{ and } \|u \| = \|v \| \right\}.$$
Let $u= (a,b,c)$, $v = (b,-a,c) \in \mathbb{Z}^3$. Then $\|u \| = \|v \|$ and $\|u \|^2 + |u \cdot v| = a^2+b^2+2c^2.$$\frac{1}{2}\|u+v \|^2 = a^2+b^2+2c^2.$
It follows that $F= E \subseteq E(\mathbb{Z})$.
In fact Now, Dickson's theorem extends toby $E(\mathbb{Z})$Legendre's three-square theorem, since Philipp Lamp shown below that $E(\mathbb{Z}) = F$$E(\mathbb{Z}) \subset F$ also (as.
Then, we have an answer to what wasextension of Dickson's theorem as Question 1 in a previous version)$E(\mathbb{Z}) = F$. 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 suggests the following questionshows that (checked for integers$E(\mathbb{N})$ contains every odd number less than $5936$)$23950$, except those in $I=\{ 5, 23, 29, 65, 167 \}$, suggesting that $E(\mathbb{N}) = F \setminus I$.
Question 2: Is it true that, for $E(\mathbb{N}) = F \setminus \{ 5, 23, 29, 65, 167 \} $$u,v \in \mathbb{N}^3$ with $\|u \| = \|v \|$, the form $\frac{1}{2} \|u+v \|^2$ covers every odd
number, except those in $\{ 5, 23, 29, 65, 167 \}$?
Application: this answer proves that the form $\| A\|^2$ covers every natural number for $A \in M_3(\mathbb{Z})$.
A positive answer to Question 2the above question would prove this result for $A \in M_3(\mathbb{N})$.
Reformulation of Question 2
Take $u=v \in \mathbb{N}^3$, then $\|u \|^2 + |u \cdot v| = 2 \|u \|^2$, so by Legendre's three-square theorem, $$2\mathbb{N} \setminus \{4^k(16n+14) \ | \ k,n \in \mathbb{N}\} \subset E(\mathbb{N}).$$
So we are reduced to prove that
$$2\mathbb{N}+1 \setminus \{ 5, 23, 29, 65, 167 \} \subset E(\mathbb{N}).$$
NowFor the convenience of the reader, as pointed out bythe answer of Philipp Lampe, if $\|u \| = \|v \|$ then(of what was $\|u \|^2 + |u \cdot v| = \|u+v \|^2/2$Question 1 in a previous version) was incorporated in the post.
Then Question 2 can be reformulated as follows:
Reformulated question: Is it true that, for $u,v \in \mathbb{N}^3$ with $\|u \| = \|v \|$, the form $\|u+v \|^2/2$ covers every odd
number, except those in $\{ 5, 23, 29, 65, 167 \}$?
ComputationComputation
sage: L=[]
....: for a1 in rangeL=cover(5070):
....sage: set([2*i+1 for a2i in range(a1+111975):
....: for a3 in range])-set(a2+1L):
....: {5, 23, 29, 65, x=a1**2+a2**2+a3**2167}
....:
Code
# %attach b=0
...SAGE/3by3.: spyx
from sage.all whileimport b<50*
cpdef andcover(int b**2<xr):
....: cdef int a1,a2,a3,b1,b2,b3,x,n
cdef list b+=1L
....: L=[]
for b1a1 in range(b+1r):
....: for a2 in bb=0
....range(a1+1):
whilefor bb<50a3 andin bb**2<x-b1**2range(a2+1):
....: bb+=1x=a1**2+a2**2+a3**2
....: for b2b1 in range(bb+1isqrt(x):
....+1): bbb=0
....: whilefor bbb<50b2 andin bbb**2<xrange(isqrt(x-b1**2-b2**2)+1):
....: bbb+=1
....: for b3 in range(bbb+1isqrt(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
....: L.append(n)
....: l=list(set(L)); l.sort()
....: s=set(range(5936))-set(l)
....: S=[]
....: for i in s:
....: if f=list(factoris_odd(i)n)
....: a=f[0][0]
....: b=f[0][1]
....: and not n ifin a<>2L:
....: S.append(i)
....: elif Integer(b).mod(2)==0:
....: S.append(i)
....: elif Integer(i/(2**b)).mod(8)<>7:
....: SL.append(i)
....: S.sort(n)
....: S
[5, 23, 29, 65,return 167]L
