Is it possible to constructively prove that every quaternion has a square root? Is it possible to constructively prove that every $q \in \mathbb H$ has some $r$ such that $r^2 = q$? The difficulty here is that $q$ might be a negative scalar, in which case there might be "too many" values of $r$. Namely, $r$ could then equal any vector quaternion of magnitude $\sqrt{|q|}$. The presence of this seemingly severe discontinuity suggests that there can't be a way to constructively prove that every quaternion has a square root.
The variety of constructivism can be as strong as possible. So any Choice principle, or Markov's Principle, or Bar Induction, is allowed.
My thoughts were to do some kind of reduction to $LPO$ or $LLPO$ or $LEM$. But I don't see how.
The way to find a square-root classically is as follows: If $q = w + xi + yj + zk$ is not a scalar quaternion, then it lies on a unique "complex plane". This is due to the fact that a vector quaternion (of the form $xi + yj + zk$) always squares to $-(x^2 + y^2 + z^2)$, which is a negative scalar. The problem then reduces to finding the square root of a complex number. The difficulty is exactly in the case when $x=y=z=0$ and $w < 0$, in which case $q$ and $r$ lie on all complex planes.
 A: ([edit] The discussion between me and Andrej refers to an earlier version of the argument, which was more confusing than this one)
The operation is not Type 2 computable. The argument is similar to how the set $\mathbb R$ is not computably equivalent to its decimal representation. This latter statement is called the tablemaker's dilemma. Constructivists and Type 2 computability theorists make use of a redundant "nega-binary" representation of real numbers instead.
Let $q=-1 + 0i + 0j + 0k$. Assume the T2TM (Type 2 Turing Machine) outputs a quaternion $r$. This $r$ is a vector. Now observe that the machine must have read only finitely many digits of the nega-binary representation of $q$. Displace $q$ by some vector $v$ which is not parallel to $r$, where the vector $v$ has magnitude smaller than $2^{-n}$, where $n$ is the number of nega-binary digits the machine has read. The machine must give the same output because the prefix of the new input is the same, but this output is wrong.
To demonstrate that two quaternions very close to $-1$ may have very different square roots: Consider $-1 + \epsilon i$: Its square roots are $\pm (i + \frac{\epsilon}2)+ o(\epsilon)$. Now consider $-1 + \delta j$: Its square roots are $\pm (j + \frac{\delta}2)+ o(\delta)$. Now the distance between each of each of these sets is at least $\sqrt{2}$, which is much greater than zero. If after reading $n$ digits of $q = -1.0 + 0i + 0j + 0k$, the machine decides to output the first digits of $0 + 1i + 0j + 0k$, then one can play a trick on it by changing $q$ to $q' = -1 + 0i + 10^{-2n}j + 0k$. Those first digits of the output will then be completely wrong.
This T2TM argument is probably a valid Type 1 argument. In which case, it provides a convincing proof that quaternion square-root is uncomputable, and therefore can not be proved constructively.
It would be nice to see a "purer" proof that reduces to LPO or some other such principle, but I can't think of one. [edit] See below.
A: Reduction to LLPO (Lesser Limited Principle of Omniscience).
The statement LLPO is the following (from Wikipedia): For any sequence a0, a1, ... such that each ai is either 0 or 1, and such that at most one ai is nonzero, the following holds: either a2i = 0 for all i, or a2i+1 = 0 for all i, where a2i and  a2i+1 are entries with even and odd index respectively. 
This is considered a quintessentially non-constructive claim.
The claim that every quaternion has a square root implies LLPO.
Consider a sequence $(p_n)_{n \geq 1} \in \{0,1\}$, with the property that at most one element of the sequence is equal to $1$. Consider the following infinite quaternionic series $q = -1 + i\left(1 - \sum_{n=1}^\infty\frac{1 - p_{2n}}{2^n}\right) + j\left(1 - \sum_{n=1}^\infty\frac{1 - p_{2n+1}}{2^n}\right)$. The series clearly converges. Now we assume that we can get an $r$ such that $r^2 = q$. Consider the angle $\theta$ between $r$ and $i$ (considered as 4d vectors with the standard inner product), and likewise consider the angle $\phi$ between $r$ and $j$. Either $\theta > \arctan(1/2)$ or $\phi > \arctan(1/2)$, as these two open regions cover all the non-zero quaternions. If $\theta > \arctan(1/2)$ then we conclude that all $p_{2n}=0$. If $\phi > \arctan(1/2)$ then we conclude that all $p_{2n+1}=0$. This is exactly LLPO.
