Quaternions and 4-vectors I recently realised that quaternions could be used to write intervals or norms of vectors in special relativity:
$$(t,ix,jy,kz)^2 = t^2 + (ix)^2 + (jy)^2 + (kz)^2 = t^2 - x^2 - y^2 - z^2$$
Is it useful? Is it used? Does it bring anything? Or is it just funny?
 A: Cornelius Lanczos has a chapter on quaternions and special relativity in his "The Variational Principles of Mechanics".  So, is has been used. But it seems more straightforward to consider the multivector algebra of spacetime so t,x,y,z really are on the same footing.
A: There is a book:
"Quaternions, Clifford Algebras and Relativistic Physics." by Patrik R. Girard. Find this if you want to learn more -- very good reading, not very complex and not very long. I'll just cite the first paragraph of chapter 3.

From the very beginning of special relativity, complex quaternions have been used
  to formulate that theory [45]. This chapter establishes the expression of the Lorentz
  group using complex quaternions and gives a few applications. Complex quaternions
  constitute a natural transition towards the Clifford algebra H ⊗ H.

Well and the reference:

[45] L. Silberstein, The Theory of Relativity, Macmillan, London, 1914.

A: You've stumbled on a fertile area.  Though not strictly what you were asking about, I can tell you that perhaps the most interesting relationship between orthogonal groups and quaternions comes from looking at spinors.  As you may know, the symmetry group called $Spin$ double covers the group of rotations, and is the more relevant group for physics since spinors transform under this larger group.  A useful example is the double cover $SU(2) \rightarrow SO(3),$ that is, $SU(2) = Spin(3)$ in the Euclidean signature.
Topologically, $SU(2)$ is a 3-sphere, which we can think of as the unit quaternions (remember, the norm is Euclidean, as pointed out by others).  To understand the map $SU(2) \rightarrow SO(3),$ let $v$ be an imaginary quaternion (which we can think of as a 3-vector), and let $q$ be in $SU(2).$  Then since multiplication of quaternions preserves the norm,
$$\overline{q} v q$$
has the same norm as v, and you will note that it is still imaginary.  Thus, the action $v\stackrel{q}{\mapsto} \overline{q}vq$ of $SU(2)$ on $R^3$ (the imaginary quaternions) is by a rotation.  Furthermore, $q$ and $-q$ act the same way.  So we have described a double cover of $SU(2)$ onto 3-dimensional rotations.
Maybe this is not what you immediately inquired about or discovered, but it's probably worth knowing.
A: First note that $q = t + xi + yj +zk$ and $q\prime = t - xi - yj - zk$
So $qq\prime = t^2 + x^2 + y^2 + z^2$
Hence, it should be $ t^2 + x^2 + y^2 +z^2$ and hence it is not the right signature of special relativity. Secondly since quaternion does not have an analog of a notion like holomorphic function (since its conjugate is not independent) it is not physically ideal or useful as far as I know. You can generalize it and have excellent usefulness but you have to give up its division algebra property. Clifford algebra is an example.
A: The object you're talking about is called, in mathematics, a Clifford algebra. The case when the algebra is over the complex field in general has a significantly different structure from the case when the algebra is over the real field, which is important in Physics. In Physics, in the specific case of 4 dimensions, using the Minkowski metric as you have in your Question, and over the complex field, the algebra is called the Dirac algebra. Once you have the name Clifford algebra, you can look them up in Google, where the first entry is, unsurprisingly, Wikipedia, http://en.wikipedia.org/wiki/Clifford_algebra, which gives you a reasonable flavor of the abstract construction methods that mathematicians prefer. The John Baez page that is linked to from the Wikipedia page is well worth reading (if you spent a year learning everything that John Baez has posted over the years, almost always with unusual clarity and engagingly, you would know most of the mathematics that might be useful for Physics).
It's not so much that the Clifford algebras are funny. Their quadratic construction is interrelated, often closely, with many other constructions in mathematics.
There are people who are enthusiastic about Clifford algebras, sometimes very or too much so, and a lot of ink has been spilled (Joel Rice's and Luboš Motl's Answers are rather inadequate to the literature, except that I think they chose to interpret your Question narrowly where I've addressed what your construction has led to in Mathematics more widely), but there are many other fish in the sea to admire.
EDIT: Particularly in light of Marek's comments below, it should be said that I interpreted Isaac's Question generously. There is a somewhat glaring mistake in the OP that is pointed out by Luboš (which I hope you see, Isaac). Nonetheless there is a type of construction that is closely related to what I chose to take to be the idea of the OP, Clifford algebras.
Isaac, this is how I think your derivation ought to go, if we just use quaternions, taking $q=t+ix+jy+kz$, 
$$q^2=(t+ix+jy+kz)(t+ix+jy+kz)=t^2-x^2-y^2-z^2+2t(ix+jy+kz).$$
The $xy,yz,zx$ terms cancel nicely, but the $tx,ty,tz$ terms don't, unless we do as Luboš did and introduce the conjugate $\overline{q}=t-ix-jy-kz$. This, however, doesn't do what I take you to be trying to do. So, instead, we introduce a fourth object, $\gamma^0$, for which $(\gamma^0)^2=+1$, and which anti-commutes with $i$,$j$, and $k$. Then the square of $\gamma^0t+ix+jy+kz$ is $t^2-x^2-y^2-z^2$. The algebra this generates, however, is more than just the quaternions, it's the Clifford algebra $C(1,3)$.
EDIT(2): Hi, Isaac. I've thought about this way too much overnight. I think now that I was mistaken, you didn't make a mistake. I think you intended your expression $(a,b,c,d)^2$ to mean the positive-definite inner product $a^2+b^2+c^2+d^2$. With this reading, however, we see three distinct structures, the positive-definite inner product, the quaternions, and the Minkowski space inner product that emerges from using the first two together. Part of what made me want to introduce a different construction is that in yours the use of the quaternions is redundant, because you'd get the same result that you found remarkable if you just used $(a,ib,ic,id)^2$ (as Luboš also mentioned). Even the positive-definite inner product is redundant, insofar as what we're really interested in is just the Minkowski space inner product. Also, of course, I know something that looks similar and that has been mathematically productive for over a century, and that can be constructed using just the idea of a non-commutative algebra and the Minkowski space inner product.
To continue the above, we can write $\gamma^1=i$, $\gamma^2=j$, $\gamma^3=k$ for the quaternionic basis elements, together with the basis element $\gamma^0$, then we can define the algebra by the products of basis elements of the algebra, $\gamma^\mu\gamma^\nu+\gamma^\nu\gamma^\mu=2g^{\mu\nu}$. Alternatively, for any vector $u=(t,x,y,z)$ we can write $\gamma(u)=\gamma^0u_0+\gamma^1u_1+\gamma^2u_2+\gamma^3u_3$, then we can define the algebra by the product for arbitrary 4-vectors, $\gamma(u)\gamma(v)+\gamma(v)\gamma(u)=2(u,v)$, where $(u,v)$ is the Minkowski space inner product. Hence, we have $[\gamma(u)]^2=(u,u)$. Now everything is getting, to my eye, and hopefully to yours, rather neat and tidy, and nicely in line with the conventional formalism.
A: It's just funny. Note that your equation doesn't actually use any single general quaternion. You only use the $i,j,k$ imaginary units in an ad hoc way to get three minus signs whenever you need them.
If you were using an actual quaternion
$$ q = t + xi + yj + zk,$$
then the only semi-natural real bilinear invariant you may construct out of it is
$$ q\bar q =  (t + xi + yj + zk ) ( t - xi - yj - zk) =  t^2+ x^2 +y^2 +z^2 $$
so the 4 real components in a quaternion still have the Euclidean, rather than Minkowskian, signature. But even for a 4-dimensional Euclidean space, the quaternions are actually just a game because we haven't really used the main nontrivial structure of the quaternions, their multiplication, in any nontrivial way. Quaternions are not genuine quaternions if you never use the relations $ij=-ji = k$ and its cyclic permutations - and we haven't used them above. We only used the fact that $i,j$ etc. anticommute with each other, but we didn't really care what their product is.
Because we haven't really used those relations, we haven't used full quaternions - except as a meaningless bookkeeping device. In the same way, one may organize 8 real numbers under the umbrella of a single "octonion" except that if the complicated and cool octonion multiplication table - with the $G_2$ automorphism group - is never employed, it's clear that the "octonion" interpretation was just a game to give a name to a collection of 8 numbers. But not every collection of 4 or 8 numbers deserves to be called "quaternion" and "octonion", even though, of course, one may get the individual components out of the "quaternion" and "octonion", too.
In the very same way, a general pair of two real numbers is simply not a complex number. By its very essence, a complex number must act as one number - so there must be a notion of holomorphy required somewhere or everywhere in the formalism - rather than two numbers. The references linked in the other answers don't understand the purpose and relevance of all those mathematical structures, so they lead to incorrect answers to the fundamental question whether the trick is real or just a fun. The right answer is that it is just a fun, and your fun even used a wrong signature that differs from a somewhat more natural fun.
A: Square any quaternion, not necessarily one involving space and time:
$$(a, b, c, d)^2 = (a^2 - b^2 - c^2 -d^2, 2ab, 2ac, 2ad)$$
The first term in invariant under a Lorentz transformation. The next three terms were omitted by the OP. Squaring a quaternion generates another quaternion. Under a Lorentz transformation, the "space-times-time" terms will change.
Now ask the converse question: what sort of physics results if the space-times-time term is invariant for two different observers? In this case, then the interval term will change. The only area of physics I know where intervals change (are dynamic) is gravity. If gravity is due to a new invariance principle in Nature (the space-times-time is the same for different observers in a gravitational field), then like special relativity, one does not have a field equation. Without a field equation, there is no force particle.
What was forgotten in the original question may be the most interesting thing to think about carefully. 
A: The construction that you've detailed is a little adhoc because you're not using the norm on the quaternions. However, there is a way of modifying the multiplication on the quaternions that allows you to do exactly that.
James Cockle introduced the split-quaternions in 1843 where he specified $i^2=-1, j^2=+1, k^2=+1$. If we modify this further by specifying that $i^2=+1$ as well then we get for a quaternion $v=a+bi+cj+dk$  (of this type) the norm $v^2=v^*v=a^2-b^2-c^2-d^2$ where $v^*=a-bi-cj-dk$ is the conjugate of $v$. This is the Minkowski metric expressed naturally through a norm. 
I don't know if there is a standard name for this type of quaternion. Coquaternions might be a possibility but it seems this is also a synonym for the split quaternions. Unfortunately, unlike the usual quaternions not all non-zero elements have an inverse. But there is a natural condition that allows you to say you can: so long as the norm of the element does not vanish then we can take the inverse. This is reminiscent of how determinants work in linear algebra. 
Minkowski space is usually expressed as a normed vector space of signature $(1,3)$. what we've just shown is that we can naturally place an algebra structure on this space. And similarly for such spaces of any signature which leads to the concept of Clifford Algebras where we specify that $p$ generators square to +1 and $q$ generators square to -1. 
How useful this is for relativity, I'm not sure. But finding extra natural structures to work with is usually a good idea.  
A: It's true that quaternion multiplication gets you the relativity space-time interval. If you could get the other three terms to mean something interesting that fit in with that, you'd have something interesting.
People tend to use quaternions only to do rotations. This involves $YXY^t$.
What do you get when you just do $YX$?
When you just do $YX$ you get a Keplerian elliptical orbit instead of a rotation.
Set $X$ to $[0,A]$ where $A$ is the vector from the center of the ellipse to the nearest orbital point from the focus, with the time parameter set to zero. (For convenience I set $|A|=1$ since the scale is arbitrary.
Set G as any unit quaternion $[0,B]$. $B\times X$ is the semiminor axis of the ellipse. $B.\overline{X}$ is the eccentricity.
$|B|X$ is the focus.
For any mean anomaly $E$, find $Y=[cos(E), sin(E)G]$ and if you multiply that by any quaternion on the orbit you will get another quaternion on the orbit rotated that far. The time dimension of that quaternion will show how far advanced or retarded the time is relative to a whole orbital period.
$YXY^t$ gives a way to remove the eccentricity when $A$ and $B$ are not perpendicular. 
Calculate a Kepler orbit in two steps. Find the mean anomaly for the input case you want, and do one quaternion multiplication.
Anything you do with quaternion multiplication when one of the quaternions is a unit vector, is analogous to calculating an elliptical orbit. 
Can you think of a way to make special relativity calculations analogous to calculating elliptical orbits?
