A pythagorean triple is a triple of integers $(a,b,c)$ with $a^2 + b^2 = c^2$. Given a triple, $(a/c, b/c)$ is a point on the unit circle, so we may associate to it the normalized angle
$$\theta\_{a,b} := \frac{\tan^{-1}(b/a)}{2\pi}\in (-1/4, 1/4)$$
Let $A$ denote the $\mathbb{Q}$-vector space spanned by $\theta\_{a,b}$ as $(a,b,c)$ ranges over all pythagorean triples.
By a theorem of Niven, $\theta\_{a,b}$ is almost always irrational, so we have that $\dim\_{\mathbb{Q}}(A)\ge 2$.


Can we do better? I would guess that the dimension is infinite, but I don't know. Maybe this has been studied somewhere?



$\mathbb{Q}$-rank of the space of angles of pythagorean triples
Yes, the dimension is infinite.


${\bf Z}$-linear relations on $\frac1{2\pi}\tan^{-1}(b/a)$
correspond to monomials in $(a+ib)/c$ that equal $1$.
But for each prime $p \equiv 1 \bmod 4$ we may choose integers $m,n$
such that $p = m^2 + n^2$ and let $(a+ib)/c = (m+in)/(m-in)$;
these are multiplicatively independent by unique factorization in ${\bf Z}[i]$.


