Every Hurwitz quaternion is associate to an integral quaternion Define Hurwitz quaternions $\mathbb{I}=\{\frac{a}{2}(1+i+j+k)+bi+cj+dk\in\mathbb{H}|a,b,c,d\in{\mathbb{Z}}\}$, i.e. quaternions with coefficients either all integers or all half-integers. How can one show that every such quaternion is associate to a quaternion with integral coefficients?
One way I found is noticing that either $a$ is even and all coefficients are already integral, or $a$ is odd and then you can multiply by $\frac{1}{2}(1\pm i\pm j\pm k)$, which is a unit in $\mathbb{I}$, with appropriately chosen signs depending on the oddity of $b,c,d$. This seems to work, but it's very annoying to check every possible case...
Is there some more intelligent solution?
 A: Define the Lipschitz and Hurwitz quaternions by the most obvious integral bases:
$$ L=\mathrm{span}\{1,\mathbf{i},\mathbf{j},\mathbf{k}\}, \\ H=\mathrm{span}\{\omega,\mathbf{i},\mathbf{j},\mathbf{k}\} $$
where $\omega=\frac{1}{2}(-1+\mathbf{i}+\mathbf{j}+\mathbf{k})=\exp\big(\frac{2\pi}{3} \frac{\mathbf{i}+\mathbf{j}+\mathbf{k}}{\sqrt{3}} \big)$ is a cube root of unity.
I think the fact in question is reasonably evident if we first develop some background on how $H$ decomposes; I think this counts as a "more intelligent" solution than brute force (even though it's more work) is because exploring how $H$ decomposes is a natural thing to do in the first place.
One decomposition that ought to be clear is $H=L\sqcup (L+\omega)$. Indeed, the sphere packing (with $H$ as the center points) is created in the first place by first doing the "obvious" $L$-centered packing and then filling in the "holes" that exist in the hypercube centers (because $\sqrt{4}=2$ just barely). Thus $[H:L]=2$.
We can also think multiplicatively. Now $H^{\times}=2T$ is the binary tetrahedral group comprised of the vertices of the $24$-cell, and $L^{\times}=Q_8$. By considering order, we know $H^{\times}/L^{\times}$ is cyclic of order $3$ generated by the cube root of unity $\omega$. This will be the key.
Also note $\omega L=L\omega$ and $\bar{\omega}L=L\bar{\omega}$ because $L^{\times}\triangleleft H^{\times}$ is normal and $L=\mathrm{span}\,L^{\times}$.
Consider the three index two sublattices $L,\omega L,\bar{\omega}L$ (note $\bar{\omega}=\omega^2=-\omega-1$) in $H$. To figure out what their intersection (or, it turns out equivalently, pairwise intersections) are, we want a nice multiplication table to find nice bases. The following one does the trick:
$$\begin{array}{rcl} \omega\mathbf{i}=\bar{\omega}+\mathbf{j}=\mathbf{k}\omega & & \bar{\omega}\mathbf{i} = -\omega+\mathbf{k}=\mathbf{j}\bar{\omega} \\ \omega\mathbf{j}=\bar{\omega}+\mathbf{k}=\mathbf{i}\omega & & \bar{\omega}\mathbf{j} = -\omega+\mathbf{i}=\mathbf{k}\bar{\omega} \\ \omega\mathbf{k} = \bar{\omega}+\mathbf{i}=\mathbf{j}\omega & & \bar{\omega}\mathbf{k}=-\omega+\mathbf{j}=\mathbf{i}\bar{\omega} \end{array} $$
You can get one equation like $\omega\mathbf{i}=\bar{\omega}+\mathbf{j}$ manually, then the rest follow by various combinations of (a) applying the automorphism of $\mathbb{H}$ which cycles $\mathbf{i},\mathbf{j},\mathbf{k}$ (and fixes $\omega,\bar{\omega}$), (b) applying quaternion conjugation, (c) conjugating by $\omega$ or $\bar{\omega}$, or (d) using the relation $\omega+\bar{\omega}=-1$.
Another nice fact is $2,2\omega,2\bar{\omega}\in L$ which implies $2\in L\cap \omega L\cap \bar{\omega} L$.
Therefore, we may multiply $L$'s standard basis by $\omega$ to get
$$ \begin{array}{ll} \omega L & =\mathrm{span}\{\omega,\omega\mathbf{i},\omega\mathbf{j},\omega\mathbf{k}\} \\ & =\mathrm{span}\{\omega,\bar{\omega}+\mathbf{i},\bar{\omega}+\mathbf{j},\bar{\omega}+\mathbf{k}\} \\ & = \mathrm{span}\{\omega,1+\mathbf{i},1+\mathbf{j},1+\mathbf{k}\}. \end{array} $$
The last equation comes from adding $\omega$ and then $2$ to the last three terms. Similar for $\bar{\omega}L$, so
$$ \begin{array}{rl} L & =\mathrm{span}\{1,1+\mathbf{i},1+\mathbf{j},1+\mathbf{k}\}, \\ \omega L & = \mathrm{span}\{\omega,1+\mathbf{i},1+\mathbf{j},1+\mathbf{k}\}, \\ \bar{\omega}L & = \mathrm{span}\{\bar{\omega},1+\mathbf{i},1+\mathbf{j},1+\mathbf{k}\} \end{array} $$
from which we conclude the intersection of any two (hence all three) sublattices is
$$ W=\mathrm{span}\{2,1+\mathbf{i},1+\mathbf{j},1+\mathbf{k}\} $$
and $H/W$ has $\{0,1,\omega,\bar{\omega}\}$ as a set of coset representatives.
So every Hurwitz integer is in one of $L,\omega L,\bar{\omega}L$, and so evidently associate to a Lipschitz integer.
A: Here's a much simpler solution. Let $w \in \mathbb{I}$ and suppose $w$ is not already integral. Look at $\bar{w}$ and choose the signs in $\epsilon = \frac{1}{2} (\pm 1 \pm i \pm j \pm k)$ such that $\bar{w} + \epsilon$ (yes, it's a plus sign, not multiply) is an integral quaternion with coefficients being even integers.
Then $N(w) + w\epsilon = w\bar{w} + w\epsilon = w(\bar{w}+\epsilon) = $ an integral quaternion. The last equality follows because $\bar{w} + \epsilon$ has even integer coefficients, so when it multiplies any Hurwitz quaternion you get an integral quarternion (simply check the expansion mentally). Subtracting $N(w)$ throughout gives the result that $w\epsilon$ is an integral quaternion.
