It occurred to me that I hadn’t written about number theory in a while. Here’s a good application of algebraic numbers, which gives a fairly simple proof of a theorem that has a very complicated elementary proof.

Let us look at primes that leave a remainder of 1 when divided by 4 (we say they’re equivalent to 1 mod 4). We have 5 = 1^2 + 2^2, 13 = 2^2 + 3^2, 17 = 1^2 + 4^2, and 29 = 2^2 + 5^2. This doesn’t always apply to composite numbers: there’s no way of expressing 15 as the sum of two squares.

Now, there’s an elementary proof of this due to Euler and a few more non-elementary ones, but they either rely on difficult concepts or are too long to remember. A better proof relies on algebraic number theory.

First, the ring **Z**[*i*] = {*a* + *bi*: *a* and *b* are in **Z**}, where *i* is the square root of -1, is a unique factorization domain, a fact I’ll prove shortly. So if a number *p* in the ring divides a product *cd* but not *c* or *d*, it’s reducible, say as *qr* where *q* and *r* are not units.

Second, we can define a function N on the ring by N(*a*+*bi*) = *a*^2 + *b*^2. Clearly, N(*a*+*bi*) will always be a positive (rational) integer, since *a* and *b* are integers. More interestingly, N is a multiplicative function – in other words, N(*xy*) = N(*x*)*N(*y*) for all *x* and *y* in the ring.

To see why that holds, let’s just do the calculation. N(*a*+*bi*)N(*c*+*di*) = (*a*^2 + *b*^2)(*c*^2 + *d*^2) = *a*^2**c*^2 + *a*^2**d*^2 + *b*^2**c*^2 + *b*^2**d*^2. On the other hand, since *i***i* = -1, we have N((*a*+*bi*)(*c*+*di*)) = N(*ac*–*bd* + (*ad*+*bc*)*i*) = (*ac*–*bd*)^2 + (*ad*+*bc*)^2 = *a*^2**c*^2 – 2*abcd* + *b*^2**d*^2 + *a*^2**d*^2 + 2*abcd* + *b*^2**c*^2 = *a*^2**c*^2 + *a*^2**d*^2 + *b*^2**c*^2 + *b*^2**d*^2.

Third, if N(*a*+*bi*) = 1, then *a*+*bi* is a unit. This is because 1 = *a*^2 + *b*^2 = (*a*+*bi*)(*a*–*bi*), so we have 1/(*a*+*bi*) = *a*–*bi* which is in the ring.

Fourth, if *p* is any prime integer, then *p* clearly doesn’t divide *x*+*i* or *x*–*i* where *x* is any integer. On the other hand, if *p* is equivalent to 1 mod 4, then we can write *p* as 4*n* + 1, where *n* is an integer. But as I will show in a moment, *p* divides *x*^2 + 1 when *x* = (2*n*)!, which means it’s not prime in the ring. Therefore, it’s reducible, say to *qr*. N(*p*) is simply *p*^2, and *qr* are not units, so N(*q*), N(*r*) > 1. Therefore, N(*q*), N(*r*) = *p*. Writing *q* as *a* + *bi*, we get *p* = *a*^2 + *b*^2.

To see that *p* divides ((2*n*)!) ^2 + 1, note that *k* is equivalent to -(4*n* + 1 – *k*) mod *p*, so (2*n*)! is equivalent to ((-1)^2*n*)(2*n*+1)(2*n*+2)…(4*n*). We have (-1)^2*n* = 1^*n* = 1, so we get ((2*n*)!)^2 = (2*n*)! * (2*n*+1)…(4*n*) = (4*n*)! mod *p*.

Now, for each number *c* between 1 and 4*n*, we can find another number *d* such that *cd* = 1 mod *p*. If we can’t, then there exist no integers *d* and *q* such that *cd* – *pq* = 1, so the ideal (*c*, *p*) in **Z** doesn’t contain 1, so it’s smaller than **Z**. But (*p*) is a maximal ideal in **Z**, so (*c*, *p*) = (*p*) and *c* = *pq* for some *q*. In particular, *c* is not in the range [1, 4*n*].

In fact, unless *c* is 1 or 4*n*, *d* will be different from *c*. Otherwise, we get *c*^2 = 1 mod *p*, so *p* divides *c*^2 – 1 = (*c*-1)(*c*+1). This means that *p* divides *c*-1 (in which case *c* = 1) or *c*+1 (in which case *c* = 4*n*). So we get 2*n*-1 pairs of numbers of the form {*c*, *d*}, *cd* = 1 mod *p*. In calculating the value of (4*n*)! mod *p*, we can ignore these pairs, since 1**x* = *x* for all *x*. Then (4*n*)! = 4*n* mod *p*, and ((2*n*)!)^2 + 1 = (4*n*)! + 1 = 4*n* + 1 = 0 mod *p*, i.e. *p* divides ((2*n*)!)^2 + 1.

Actually, there’s an indirect way of showing that if *p* = 4*n* + 1 then we can always find some *x* such that *x*^2 = -1 mod *p*, but it requires me to talk a little bit about group theory and I don’t want to do that yet.