The free particle Schródinger equation is not relativistic:
$$ - \frac{\hbar^2}{2m} \nabla^2 \phi= i \hbar \frac{\partial}{\partial t}\phi $$
Dirac had to figure out a way to make a Lorentz-invariant theory.
\begin{eqnarray*}
E^2 &=& (mc^2)^2 + (pc)^2 \\
\left( - \frac{1}{c^2 }\frac{\partial^2 }{\partial t^2} + \nabla^2 \right)\phi &=& \frac{m^2 c^2 }{\hbar^2 }\phi^2
\end{eqnarray*}
At this point we could set $c = \hbar = 1$ which are called "natural units"
$$ ( \square- m) \phi = 0 $$
The probability current density is no longer positive-definite.
$$ ( i\partial - m)\phi = 0 $$
In textbooks these will be solved over a flat space like $\mathbb{R}^4$ with a unique spin-structure. However, many physical theories are "compactified" to smaller spaces such as $\mathbb{R}^3 \times S^1$ it would not be uncommon to solve a theory on a compact space such as $S^1 \times S^1 \times S^1 \times S^1$ and let the circle radii vary.
If we set $e(x^\mu) = e^{2\pi i \langle x^\mu , m\rangle}$ then the $m = 0$ solution would need $m_0^2 = m_1^2 + m_2^2 + m_3^2$ and the torus geometry forces $m_0, m_1, m_2, m_3 \in \mathbb{Z}$. This is a diophantine restriction.
Also discussed on physics stackexchange:
While I'm at it, we should solve the pythagorean quadriple equations $a^2 + b^2 + c^2 = d^2$. It has solutions parameterized by four integers: \begin{eqnarray*} a &=& m^2 + n^2 - p^2 - q^2 \\ b &=& 2(mq + np)\\ c &=& 2(nq - mp)\\ d &=& m^2 + n^2 + p^2 + q^2 \end{eqnarray*} Another way to look at it is to organize into three matrices: $$ \frac{1}{d}\left[ \begin{array}{cccc} m^2 + n^2 - p^2 - q^2 & 2(np - mq) & 2(mp + nq) \\ 2(mq + np) & m^2 - n^2 + p^2 - q^2 & 2(pq - mn) \\ 2(nq - mp) & 2(mn + pq) & m^2 - n^2 - p^2 + q^2\end{array} \right] \in SO(3, \mathbb{Q} ) $$
It's known there's relationship between pythagorean triples and spinor groups. However, different communities might call different things a "spinor".