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): $$E(\mathbb{A}) = \{\|u \|^2 + |u \cdot v| \text{ such that } u,v \in \mathbb{A}^3 \text{ and } \|u \| = \|v \|\}.$$

Let $u= (a,b,c)$, $v = (b,-a,c) \in \mathbb{Z}^3$. Then $\|u \| = \|v \|$ and $\|u \|^2 + |u \cdot v| = a^2+b^2+2c^2.$ It follows that $E \subseteq E(\mathbb{Z})$. But surprisingly, the computation below suggests that $E(\mathbb{Z})=F$ also, which would be an extension of Dickson's theorem.

Question 1: Is it true that $E(\mathbb{Z})=F$?

The computation suggests also the following question:

Question 2: Is it true that $E(\mathbb{N}) = F \setminus \{ 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 2 would prove this result for $A \in M_3(\mathbb{N})$.

Computation

sage: L=[]
....: for a1 in range(-12,12):
....:     for a2 in range(-12,12):
....:         for a3 in range(-12,12):
....:             for b1 in range(-12,12):
....:                 for b2 in range(-12,12):
....:                     for b3 in range(-12,12):
....:                         if a1**2+a2**2+a3**2==b1**2+b2**2+b3**2:
....:                             n=a1**2+a2**2+a3**2+a1*b1+a2*b2+a3*b3
....:                             L.append(n)
....: l=list(set(L))
....: l.sort()
....: set(range(255))-set(l)
....:
{14, 30, 46, 56, 62, 78, 94, 110, 120, 126, 142, 158, 174, 184, 190, 206, 222, 224, 238, 248, 254}

The research of the above sequence in OEIS provides exactly one result:
A055039 Numbers of the form 2^(2i+1)*(8*j+7).

sage: L=[]
....: for a1 in range(25):
....:     for a2 in range(25):
....:         for a3 in range(25):
....:             for b1 in range(25):
....:                 for b2 in range(25):
....:                     for b3 in range(25):
....:                         if a1**2+a2**2+a3**2==b1**2+b2**2+b3**2:
....:                             n=a1**2+a2**2+a3**2+a1*b1+a2*b2+a3*b3
....:                             L.append(n)
....: l=list(set(L))
....: l.sort()
....: s=set(range(1484))-set(l)
....: S=[]
....: for i in s:
....:     f=list(factor(i))
....:     a=f[0][0]
....:     b=f[0][1]
....:     if a<>2:
....:         S.append(i)
....:     elif Integer(b).mod(2)==0:
....:         S.append(i)
....:     elif Integer(i/(2**b)).mod(8)<>7:
....:         S.append(i)
....: S.sort()
....: S
....:
[5, 23, 29, 65, 167]