I don’t know the answer to your question, but I would like to clarify this statement. The difference between two points would be merely the subtraction of their coordinates without the squares. If you’re asking for the distance between points, then what you have is almost correct, except you must take the square root of the entire argument. Correct me if I’m talking about something different than what you mean.

I’ll ponder this more after I sign off, and other Dopers may have different takes on the subject, but I suspect you’re not going to get much more of an answer than the following: that’s just the way things are. This minus sign is just one of the axioms (in fact, probably the axiom) on which all of special relativity rests (in math language, the axiom is that the symmetry group of spacetime is locally SO(3,1) plus translations), and all other consequences are derived from this axiom. You can’t ask “why” this is the case any more than you can ask “why” there exists a straight line between any two points in Euclidean geometry.

The essential point is why x[sup]2[/sup]+y[sup]2[/sup]+z[sup]2[/sup] is useful for three dimensional space. This quantity is preserved under rotations of the coordinate system. If you and I stand at the same point and measure the distance to another point, but I turn my point of view (and thus my x, y, and z axes) relative to yours, we’ll get different values for the x, y, and z coordinates. The value of x[sup]2[/sup]+y[sup]2[/sup]+z[sup]2[/sup] will remain the same, though.

Now suppose we’re both facing the same direction and we both look from the same reference point to the same test point, but I’m moving at a constant velocity passing through the point while you stand still. Relativity says that we’ll measure different values for x, y, z, and t than each other, but the value of x[sup]2[/sup]+y[sup]2[/sup]+z[sup]2[/sup]-t[sup]2[/sup] we get will be the same. The fact that this quantity (the “spacetime interval”) is preserved under this sort of “generalized rotation” while x[sup]2[/sup]+y[sup]2[/sup]+z[sup]2[/sup]+t[sup]2[/sup] isn’t has been established in many experiments.

For what it’s worth, in 4-dimensional Euclidean geometry, the square of the distance between two points is (W - w)[sup]2[/sup] + (X - x)[sup]2[/sup] + (Y - y)[sup]2[/sup] + (Z - z)[sup]2[/sup]. But space isn’t Euclidean, so that’s not necessarily the right metric.

Just to expand on what ultrafilter said, it’s perfectly possible (mathematically speaking) to have a Euclidean 4-space, whose metric is the usual 4-dimensional version of the Euclidean metric:
ds[sup]2[/sup] = dx[sup]2[/sup] + dy[sup]2[/sup] + dz[sup]2[/sup] + dw[sup]2[/sup]
(Here I arbitrarily chose to call the fourth dimension w, because in this case there’s no reason to equate it with time.)

(By the way, when I say “metric” I basically mean the formula we use for computing “distances” in that space, although “interval” is probably a more appropriate word than “distance” in the case where they can be negative.)

However, whether such a space is useful in describing the physical world is another question. Special relativity offers 4-dimensional Minkowski spacetime as a better description of the world. (In general relativity this space becomes a curved pseudo-Riemannian space because of the effect of gravity, but as far as I know – I’m still rather new to GR – it can still be approximated by flat Minkowski space at any particular point. Except, I assume, for singularities.)

Minkowski space is a pseudo-Euclidean space, in the sense that it has the same topological structure and curvature (constant zero curvature) as Euclidean space, but has a different metric, namely:
ds[sup]2[/sup] = dx[sup]2[/sup] + dy[sup]2[/sup] + dz[sup]2[/sup] - c *dt[sup]2[/sup]
where c is the speed of light, but physicists (particle physicists, at least) typically choose their units so that c = 1.

Some people like to say that time is imaginary (in which case you’d use a plus sign instead of the minus sign, and have the minus contributed by the factor of i[sup]2[/sup]). But typically physicists just use the minus sign. After all, Minkowski spacetime isn’t Euclidean 4-space, so why should its metric have the same formula?

As far as why we want a minus sign, see Mathocist’s post above. To reiterate, we want the interval to be invariant under rotations in the space. I think MikeS is trying to answer a deeper question, namely “Why do we live in a universe where intervals of the form x[sup]2[/sup] + y[sup]2[/sup] + z[sup]2[/sup] - t[sup]2[/sup] are invariant?” I’m not sure it’s even possible to answer that question.