This is the second question in my series of problems for other enthusiasts. The first is here. My apologies, as usual, if you find them insultingly trivial.
The Lipschitz quaternions are quaternions with integer coefficients. We see they form a subring of the real quaternions without closure under division. Upon inspection, it's easy to see the coefficients of the quaternions can be replaced with elements of any ring. Specifically, let's consider the quaternions whose coefficients are members of $\mathbb{Z}_n$. These are begging to be called the modular Lipshitz quaternions. Suppose we denote them $Q_n$. The following questions are natural:
(1) For which $n$, if any, does $Q_n$ have at least one proper nontrivial ideal?
Answer: the remarks of @MatheiBoulomenos and @rschwieb have shown that $Q_n$ has at least one proper, nontrivial ideal for all integers $n >1$. Wow!
(2) If $Q_n$ has proper nontrivial ideals what are their orders? How, if at all, are they related to the ideals of $\mathbb{Z}_n$?
Partial Answer: @rschwieb has pointed out that If $I$ is an ideal of $\mathbb{Z}_n$, then $Q_n$ has an ideal consisting of quaternions whose coefficients are members of $I$.
For prime $n$, (1) reduces to deciding whether or not $Q_n$ is a domain. Since any finite domain is a field, showing this would settle (1) in the negative. Note that $\mathbb{Z}_n$ is always a subring of $Q_n$, thus $Q_n$ cannot be a domain if $n$ is composite. It's also true (though possibly useless) that $Q_n$ is order $n^4$.
Edit:
Concerning the case where $n$ is prime, checking whether or not $Q_n$ is a domain may actually be nontrivial. Suppose $q$ and $p$ are two members of $Q_n$, both different from zero, and that $qp = 0$. Writing $q = (a,b,d,c)$ and $p = (e,f,g,h)$, one may check that this condition is equivalent to the matrix
$$ M_q = \begin{pmatrix} a & -b & -c & -d\\ b & a & d & -c\\ c & -d & a & b\\ d & c & -b & a\\ \end{pmatrix} $$
having a nontrivial kernel over $\mathbb{Z}_n$. Evidently $Q_n$ is a domain if and only if $\ker M_q$ is trivial for every $q \in Q_n$. Equivalently, we may state this by saying the determinant of $M_q$ must never vanish on $Q_n$. For small $n$ it is possible to check this claim exhaustively. For instance, $n = 2$ requires only $2^4 = 16$ calculations (however, rschwieb has just shown $Q_2$ fails to be a domain!).
Edit (5/1/2017) Evidently I made an error regarding the it being enough to show prime $Q_p$ are domains. An updated post, with a proposed solution and new questions, can be found here.
It's clear that if $I\lhd \mathbb Z_n$, then $\{a+bi+cj+dk\mid a,b,c,d\in I\}$ is an ideal of $Q_n$. So, it is necessary for $n$ to be prime for the $Q_n$ to be simple.
For $n=2$, $Q_2$ is not a domain, either. This is because $i^2=-1=1$ implies $(i-1)(i-1)=0$. I'm working on more information for other primes...