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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
Sebastien Palcoux
  • 5,797

1 Answers1

5

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 $$\Vert A \Vert^2 = \frac{1}{2} \left( \sum_{i=1}^3 (a_i^2 +b_i^2) + \sqrt{\left(\sum_{i=1}^3 (a_i^2 +b_i^2)\right)^2 -4\sum_{i<j}(a_ib_j-a_jb_i)^2} \right)$$

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 $\theta = \arccos(1/6) \simeq 1.403348 \text{ rad} \simeq 80.4°$

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Sebastien Palcoux
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