# Integers which are squared norm of 3 by 3 integer matrices

Let $$n$$ be a positive integer. Let $$E_n$$ be the set of integers which are the sum of $$n$$ squares.
Let $$F_n$$ be the set of integers of the form $$\Vert A \Vert^2$$ with $$A \in M_n(\mathbb{Z})$$. Then $$E_n \subseteq F_n$$ because: $$\left\| \pmatrix{a_1&0& \cdots\\ \vdots & \vdots& \\ a_n&0& \cdots} \right\|^2 = \sum_{i=1}^n a_i^2.$$ Note that the case $$n=3$$ is exceptional, because $$E_n= F_n$$ $$\forall n \neq 3$$, whereas $$E_3 \subsetneq F_3$$:

• obviously $$E_1=F_1$$,
• it is proved here that $$E_2=F_2$$,
• for $$n \ge 4$$, $$E_n=F_n$$ because $$E_4 = \mathbb{N}$$, by Lagrange's four square theorem,
• finally, $$E_3 \subsetneq F_3$$ because $$\forall n \le 2000$$, $$n \in F_3$$ (by computation below), whereas:

Legendre's three-square theorem
A natural number can be represented as the sum of three squares of integers if and only if it is not of the form $$4^n(8m+7)$$ for integers $$n,m \ge 0$$.

Question: Which integers are contained in $$F_3$$?

The computation suggests that $$F_3$$ contains every natural number, so (if it is true) we are reduced to prove that it contains those of the form $$4^n(8m+7)$$, by Legendre's three-square theorem.

Computation

sage: L=[]
....: for a2 in range(33):
....:     for a4 in range(33):
....:         for a5 in range(33):
....:             for a7 in range(33):
....:                 for a8 in range(33):
....:                     n=numerical_approx(matrix([[0,a2,0],[a4,a5,0],[a7,a8,0]]).norm()^2,digits=10)
....:                     if n.is_integer():
....:                         L.append(int(n))
....: l=list(set(L))
....: l.sort()
....: l[2095]
....:
2095


Short version

$$\left\| \pmatrix{a&0&0\\ b&0&0 \\ c&0&0} \right\|^2 = a^2+b^2+c^2 \ \text{ and } \ \left\| \pmatrix{a&a&0\\ b&-c&0 \\ c&b&0} \right\|^2 = 2a^2+b^2+c^2.$$
These two forms together cover every natural number by Theorems I and V in this paper of L.E. Dickson.

Long version

Recall that $$\Vert A \Vert^2$$ is just the largest eigenvalue of $$A^*A$$. Take $$a_i,b_i \in \mathbb{Z}$$ and $$A=\pmatrix{a_1&b_1&0\\ a_2&b_2&0 \\ a_3&b_3&0}$$ Then $$A^*A = \pmatrix{a_1^2 + a_2^2 + a_3^2&a_1b_1 + a_2b_2 + a_3b_3&0\\ a_1b_1 + a_2b_2 + a_3b_3&b_1^2 + b_2^2 + b_3^2&0 \\ 0&0&0}$$

We deduce its characteristic polynomial and the largest root. It follows that $$\Vert A \Vert^2 = \frac{1}{2} \left( \sum_{i=1}^3 (a_i^2 +b_i^2) + \sqrt{\left(\sum_{i=1}^3 (a_i^2 +b_i^2)\right)^2 -4\sum_{i

Let $$u=\pmatrix{a_1\\ a_2 \\ a_3}$$, $$v=\pmatrix{b_1\\ b_2 \\ b_3}$$, and $$u\times v$$ be their cross product. Then, observe that

$$\Vert A \Vert^2 = \frac{1}{2} \left( \Vert u \Vert^2 + \Vert v \Vert^2 + \sqrt{\left(\Vert u \Vert^2 + \Vert v \Vert^2\right)^2 -4 \Vert u \times v\Vert^2} \right)$$

Recall that $$\Vert u \times v \Vert^2 + (u \cdot v)^2 = \Vert u \Vert^2\Vert v \Vert^2$$, with $$u \cdot v$$ the dot product. Then

$$\Vert A \Vert^2 = \frac{1}{2} \left( \Vert u \Vert^2 + \Vert v \Vert^2 + \sqrt{\left(\Vert u \Vert^2 - \Vert v \Vert^2\right)^2 +4 (u \cdot v)^2} \right)$$

Assume that $$\Vert u \Vert = \Vert v \Vert$$. Then $$\Vert A \Vert^2 = \Vert u \Vert^2+ \vert u \cdot v \vert.$$

For any $$u= \pmatrix{a\\ b \\ c} \in \mathbb{Z}^3$$, take $$v= \pmatrix{a\\ -c \\ b}$$. Then $$\Vert A \Vert^2 = 2a^2+b^2+c^2.$$

By Theorem V in this paper of L.E. Dickson, the above form represents every natural numbers not of the form $$2^{2n+1}(8m+7)$$. But this last is in $$E_3$$ by Legendre's three-square theorem, and we already know that $$E_3 \subset F_3$$. The result follows. $$\square$$

Bonus problem: Find a proof with $$A \in M_3(\mathbb{N})$$.

For fun: a classification of the natural numbers by angles

Recall that $$\Vert u \times v\Vert^2 = \Vert u \Vert^2 \Vert v \Vert^2 \sin^2(u,v)$$. Let's recall the above matrix $$A$$ as $$A_{u,v}$$. Consider the angle $$\alpha(n):=\min_{u,v \in \mathbb{Z}^3}\{\text{angle}(u,v) \in [0,2\pi) \text{ such that } \Vert A_{u,v} \Vert^2 = n \}.$$

Theorem: $$\alpha(n) = 0$$ if and only if $$n \in E_2E_3$$.
proof: Note that $$\alpha(n) = 0$$ iff $$\exists u,v \in \mathbb{Z}^3$$ with $$\Vert A_{u,v} \Vert^2 = n$$ and $$u \times v = 0$$ (i.e. collinear), iff $$\exists r \in \frac{1}{\gcd(u)}\mathbb{Z}$$ such that $$v=ru$$, with $$\gcd(u)$$ the greatest common divisor of $$u_1, u_2$$ and $$u_3$$. Then $$\Vert A \Vert^2= (r^2+1)\Vert u \Vert^2.$$

For any $$u'= \pmatrix{a\\ b \\ c} \in \mathbb{Z}^3$$ and any $$s,t \in \mathbb{Z}$$, assume that $$u=su'$$ (so that $$s | \gcd(u)$$) and $$r=t/s$$. Then $$\Vert A \Vert^2= (t^2+s^2)\Vert u' \Vert^2 = (t^2+s^2)(a^2+b^2+c^2).$$
The result follows. $$\square$$

By above material with $$\vert u \cdot v \vert^2 = \Vert u \Vert^2 \Vert v \Vert^2 \cos^2(u,v)$$, we have:
Lemma: $$\alpha(2a^2+b^2+c^2) \le \arccos(\frac{a^2}{a^2+b^2+c^2})$$.

Then, $$\alpha(7) \in (0,\theta]$$, with $$\theta = \arccos(1/6) \simeq 1.403348 \text{ rad} \simeq 80.4°$$

• Congratulations, this is nicely done. – user1551 Oct 24 '18 at 7:08