Integers which are squared norm of 2 by 2 integer matrices

Question: Which integers are of the form $$\Vert A \Vert^2$$, with $$A \in M_2(\mathbb{Z})$$.

The code below provides the first such integers: $$0, 1, 2, 4, 5, 8, 9, 10, 13, 16, 17, 18, 20, 25, 26$$.
By searching this sequence on OEIS, we find: "Numbers that are the sum of 2 squares" A001481.

Are these integers exactly those which are the sum of two squares ?

Research

First, $$\Vert A \Vert^2$$ is the largest eigenvalue of $$A^*A$$, so for $$A = \left( \begin{matrix} a & b \cr c & d \end{matrix} \right)$$ and $$a,b,c,d \in \mathbb{Z}$$, so we get:
$$\Vert A \Vert^2 = \frac{1}{2} \left(a^2+b^2+c^2+d^2+\sqrt{(a^2+b^2+c^2+d^2)^2 - 4(ad-bc)^2}\right)$$

Obviously, every sum of two squares is of the expected form , because by taking $$c=d=0$$, we get $$\Vert A \Vert^2=a^2+b^2$$.

Then it remains to prove that there is no other integer (if true).

Now, recall that:

Sum of two square theorem
An integer greater than one can be written as a sum of two squares if and only if its prime decomposition contains no prime congruent to 3 (mod 4) raised to an odd power.

By taking $$c=ra$$ and $$d=rb$$, we get that $$\Vert A \Vert^2 = (r^2+1)(a^2+b^2)$$, which is also a sum of two square because the following equation occurs (proof here):
$$r^2 \not \equiv -1 \mod 4s+3$$

A necessary condition for $$\Vert A \Vert^2$$ to be an integer, is that $$(a^2+b^2+c^2+d^2)^2 - 4(ad-bc)^2$$ must be a square $$X^2$$, so that $$(X,2(ad-bc),a^2+b^2+c^2+d^2)$$ is a Pythagorean triple, so must be of the form $$(k(m^2-n^2),2kmn,k(m^2+n^2)$$, and then $$\Vert A \Vert^2 = km^2$$. So it remains to prove that $$k$$ must be a sum of two squares.

sage: L=[]
....: for a in range(-6,6):
....:     for b in range(-6,6):
....:         for c in range(-6,6):
....:             for d in range(-6,6):
....:                 n=numerical_approx(matrix([[a,b],[c,d]]).norm()^2,digits=10)
....:                 if n.is_integer():
....:                     L.append(int(n))
....: l=list(set(L))
....: l.sort()
....: l[:20]
....:
[0, 1, 2, 4, 5, 8, 9, 10, 13, 16, 17, 18, 20, 25, 26, 29, 32, 34, 36, 37]


Yes. If $$A$$ is a $$2\times2$$ integer matrix such that $$n=\|A\|^2$$ is an integer, $$n$$ must be the sum of two integer squares. Conversely, if $$n$$ is the sum of two integer squares, then $$n=\|A\|^2$$ for some $$2\times2$$ integer matrix $$A$$.
Proof. Suppose $$A$$ is a $$2\times2$$ integer matrix such that $$n=\|A\|^2$$ is an integer. We want to show that $$n$$ is the sum of two integer squares. This is clearly true if $$n$$ is $$0$$ or $$1$$. Suppose $$n>1$$. Then $$A^TA-nI$$ is a singular matrix with integer entries. Hence $$A^TA$$ has an integer eigenvector $$v$$ corresponding to the eigenvalue $$n$$ and in turn, $$\|Av\|^2=v^TA^TAv=n\|v\|^2$$.
Since both $$\pmatrix{x\\ y}:=v$$ and $$\pmatrix{a\\ b}:=Av$$ are integer vectors, the previous equality implies that $$n(x^2+y^2)=a^2+b^2$$. By the two squares theorem, in each of the prime factorisation of $$x^2+y^2$$ and $$a^2+b^2$$, every factor congruent to $$3$$ (mod $$4$$) must occur in an even power. Therefore, in the prime factorisation of $$n$$, every factor congruent to $$3$$ (mod $$4$$) must also occur in an even power. Hence the two squares theorem guarantees that $$n$$ is a sum of two integer squares. This proves one direction of our assertion.
For the other direction, suppose $$n=a^2+b^2$$ for some two integers $$a$$ and $$b$$. Then $$\|A\|^2=n$$ when $$A=\pmatrix{a&-b\\ b&a}$$ or $$\pmatrix{a&0\\ b&0}$$.
• I see. Then note that this theorem is stated with $n > 1$, but it is ok. Oct 20, 2018 at 17:36