Theorems V in [this paper][1] of [L.E. Dickson][2] 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][5], $E(\mathbb{Z}) \subset F$ 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 $23950$, 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][4] 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(70) sage: set([2*i+1 for i in range(11975)])-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 [1]: https://mathscinet.ams.org/mathscinet-getitem?mr=1561323 [2]: https://en.wikipedia.org/wiki/Leonard_Eugene_Dickson [3]: https://en.wikipedia.org/wiki/Dot_product [4]: https://math.stackexchange.com/a/2967989/84284 [5]: https://en.wikipedia.org/wiki/Legendre%27s_three-square_theorem