First of all, what do I mean by a three-dimensional number system? We say that the complex numbers are two-dimensional, and that the quaternions are four-dimensional, but what do we mean by these things? There's a few potential answers to this question, but for our purposes we'll take the following narrative:

Complex numbers can be written in the form (a+bi), where a and b are real numbers. For the variable-averse, this just means we have things like (3+6i) and (5-2i) and (-8+3i). Some amount of "units" (that is, ones), and some amount of i's.

Most people are happy to stop here and say "well, there's two numbers that you're using, so that's two dimensions, ho hum". I think that's underselling it, though, since there's something nontrivial and super cool happening here. See, each complex number has an "absolute value", which is its distance from zero. If you imagine "3+6i" to mean "three meters East and six meters North", then the distance to that point will be 6.708 meters. We say the absolute value of (3+6i), which is written like |3+6i|, is equal to 6.708. Similarly, interpreting "5-2i" to mean "five meters East and two meters South" we get that |5-2i| = 5.385.

The neat thing about this is that absolute values multiply really nicely. For example, the two numbers above multiply to give (3+6i) × (5-2i) = (27+24i) which has a length of 36.124. What's impressive is that this length is the product of our original lengths: 36.124 = 6.708 × 5.385. (Okay technically this is not true due to rounding but for the full values it is true.)

This is what we're going to say is necessary to for a number system to accurately represent a space. You need the numbers to have lengths corresponding to actual lengths in space, and you need those lengths to be "multiplicative", which just means it does the thing we just saw. (That is, when you multiply two numbers, their lengths are multiplied as well.)

There's still of course the question of what "actual lengths in space" means, but we can just use the usual Euclidean method of measurement. So, |3+6i| = √(3²+6²) and |5-2i| = √(5²+2²). This extends directly to the quaternions, which are written as (a+bi+cj+dk) for real numbers a, b, c, d. (Don't worry about what j and k mean if you don't know; it turns out not to really matter here.) The length of the quaternion 4+3i-7j+4k can be calculated like |4+3i-7j+4k| = √(4²+3²+7²+4²) = 9.486 and similarly for other points in "four-dimensional space". These are the kinds of number systems we're looking for.

[To be explicit, for those who know the words: What we are looking for is a vector algebra over the real numbers with a prescribed basis under which the Euclidean norm is multiplicative and the integer lattice forms a subring.]

Part Two: Sums of Squares

Now for something completely different. Have you ever thought about which numbers are the sum of two perfect squares? Thirteen works, for example, since 13 = 3² + 2². So does thirty-two, since 32 = 4² + 4². The squares themselves also work, since zero exists: 49 = 7² + 0². But there are some numbers, like three and six, which can't be written as a sum of two squares no matter how hard you try. (It's pretty easy to check this yourself; there aren't too many possibilities.)

Are there any patterns to which numbers are a sum of two squares and which are not? Yeah, loads. We're going to look at a particularly interesting one: Let's say a number is "S2" if it's a sum of two squares. (This thing where you just kinda invent new terminology for your situation is common in math. "S2" should be thought of as an adjective, like "orange" or "alphabetical".) Then here's the neat thing: If two numbers are S2 then their product is S2 as well.

Let's see a few small examples. We have 2 = 1² + 1², so we say that 2 is S2. Similarly 4 = 2² + 0² is S2. Then 2 × 4, that is to say, 8, should be S2 as well. Indeed, 8 = 2² + 2².

Another, slightly less trivial example. We've seen that 13 and 32 are both S2. Then their product, 416, should also be S2. Lo and behold, 416 = 20² + 4², so indeed it is S2.

How do we know this will always work? The simplest way, as long as you've already internalized the bit from Part 1 about absolute values, is to think about the norms of complex numbers. A norm is, quite simply, the square of the corresponding distance. (Okay yes it can also mean different things in other contexts, but for our purposes that's what a norm is.) The norm is written with double bars, so ‖3+6i‖ = 45 and ‖5-2i‖ = 29 and ‖4+3i-7j+4k‖ = 90.

One thing to notice is that if your starting numbers are whole numbers then the norm will also be a whole number. In fact, because of how we've defined lengths, the norm is just the sum of the squares of the real-number bits. So, any S2 number can be turned into a norm of a complex number: 13 can be written as ‖3+2i‖, 32 can be written as ‖4+4i‖, and 49 can be written as ‖7+0i‖.

The other thing to notice is that, since the absolute value is multiplicative, the norm is also multiplicative. That is to say, for example, ‖(3+6i) × (5-2i)‖ = ‖3+6i‖ × ‖5-2i‖. It's pretty simple to prove that this will work with any numbers you choose.

But lo, gaze upon what happens when we combine these two facts together! Consider the two S2 values 13 and 32 from before. Because of the first fact, we can write the product 13 × 32 in terms of norms: 13 × 32 = ‖3+2i‖ × ‖4+4i‖. So far so good. Then, using the second fact, we can pull the product into the norms: ‖3+2i‖ × ‖4+4i‖ = ‖(3+2i) × (4+4i)‖. Huzzah! Now, if we write out the multiplication as (3+2i) × (4+4i) = (4+20i), we can get a more natural looking norm equation: ‖3+2i‖ × ‖4+4i‖ = ‖4+20i‖ and finally, all we need to do is evaluate the norms to get our product! (3² + 2²) × (4² + 4²) = (4² + 20²)

The cool thing is that this works no matter what your starting numbers are. 218 = 13² + 7² and 292 = 16² + 6², so we can follow the chain to get 218 × 292 = ‖13+7i‖ × ‖16+6i‖ = ‖(13+7i) × (16+6i)‖ = ‖166+190i‖ = 166² + 190² and indeed you can check that both extremes are equal to 63,656. No matter which two S2 numbers you start with, if you know the squares that make them up, you can use this process to find squares that add to their product. That is to say, the product of two S2 numbers is S2.

Part Four: Why do we skip three?

Now we have all the ingredients we need for our cute little proof soup! First, let's hop to the quaternions and their norm. As you should hopefully remember, quaternions have four terms (some number of units, some number of i's, some number of j's, and some number of k's), so a quaternion norm will be a sum of four squares. For example, ‖4+3i-7j+4k‖ = 90 means 90 = 4² + 3² + 7² + 4².

Since we referred to sums of two squares as S2, let's say the sums of four squares are S4. 90 is S4 because it can be written as we did above. Similarly, 7 is S4 because 7 = 2² + 1² + 1² + 1², and 22 is S4 because 22 = 4² + 2² + 1² + 1². We are of course still allowed to use zeros; 6 = 2² + 1² + 1² + 0² is S4, as is our friend 13 = 3² + 2² + 0² + 0².

The same fact from the S2 numbers still applies here: since 7 is S4 and 6 is S4, we know that 42 (the product of 7 and 6) is S4. Indeed, after a bit of fiddling I've found that 42 = 6² + 2² + 1² + 1². I don't need to do that fiddling, however, if I happen to be able to calculate quaternions! All I need to do is follow the chain, just like before: 7 × 6 = ‖2+i+j+k‖ × ‖2+i+j‖ = ‖(2+i+j+k) × (2+i+j)‖ = ‖2+3i+5j+2k‖ = 2² + 3² + 5² + 2². This is a different solution than the one I found earlier, but that's fine! As long as there's even one solution, 42 will be S4. Using the same logic, it should be clear that the product of any two S4 numbers is an S4 number.

Now, what goes wrong with three dimensions? Well, as you might have guessed, it has to do with S3 numbers, that is, numbers which can be written as a sum of three squares. If we had any three-dimensional number system, we'd be able to use the strategy we're now familiar with to prove that any product of S3 numbers is an S3 number. This would be fine, except, well…

3 × 5 = 15.

Why is this bad? See, 3 = 1² + 1² + 1² and 5 = 2² + 1² + 0², so both 3 and 5 are S3. However, you can check without too much trouble that 15 is not S3; no matter how hard you try, you can't write 15 as a sum of three squares.