Denoting {a + b sqrt(-5) for a, b integers} by Z[sqrt(-5)], it's worth pointing out that you can actually use the norm to get a bit of the FTA--namely that ever

Denoting {a + b sqrt(-5) for a, b integers} by Z[sqrt(-5)], it's worth pointing out that you can actually use the norm to get a bit of the FTA--namely that every nonzero nonunit factors into irreducibles. By x being "irreducible", I mean that x is a nonzero nonunit element of Z[sqrt(-5)] and whenever x = yz for y,z in Z[sqrt(-5)] then either y or z is a unit. So, denoting x = a + b sqrt(-5) in Z[sqrt(-5)], we define the norm[1] of x as follows: N(x) = a^2 + 5b^2 Note that, in this case, N is a function from Z[sqrt(-5)] to the natural numbers. For x, y in Z[sqrt(-5)], it turns out that the following properties of the norm hold (proving them is fairly straightforward, as it's essentially "plug-and-chug" combined with a bit of reasoning about how things work in the natural numbers): 1. N(xy) = N(x)N(y) 2. N(x) = 0 if and only if x = 0 3. N(x) = 1 if and only if x = 1 or -1 (ie x is a unit). So, if x is a nonzero nonunit that isn't irreducible, then we can, by definition, write x = y*z where y and z are also nonzero nonunits. Applying strong induction via the norm, we can show that x can be written as a product of irreducibles. Of course this product is, in general, not uniquely determined. [1]: https://en.wikipedia.org/wiki/Field_norm https://en.wikipedia.org/wiki/Field_norm