I promised that I’d explain later the crucial complication with the proof that 26 is the only integer sandwiched between a square and a cube. In fact, that complication permeates the entire discipline, and the more advanced methods it uses seem to largely be about getting around it.

First, there’s the complication is that I ignored units. In a ring with 1, a unit is any element that divides 1. The element 1 is always a unit, as is -1. Similarly, two elements that divide each other, or, if you will, have a unit quotient, are called associates. It’s important with squares, because for example -4 is associated to a square, but isn’t a square itself.

In the case that uses SQRT(-2), it’s not a problem, because the only units are +/-1, are -1 is a cube. But the ring that uses SQRT(2) has infinitely many units: 1, -1, and all the multiples and quotients resulting from SQRT(2) – 1, since (SQRT(2) – 1)(SQRT(2) + 1) = 1. In fact, the degenerate solution to the equation, x = +/-1, y = -1, is based on factoring 1^2 – 2 = (-1)^3 as (1 + SQRT(2))(1 – SQRT(2)) = (-1)^3.

Although in this particular equation this is the critical complication, in fact there’s an even bigger one, which in this case fortunately works out: I didn’t prove unique factorization.

Rational integers can be factored uniquely, up to reordering and multiplying by units: 6 = 2*3 = 3*2 = (-2)(-3). No list of primes other than one associate of 2 and one associate of 3 can generate 6. You probably learned this in elementary school, though obviously you didn’t get to see any proof.

In fact, some extensions of the integers can similarly be uniquely factored, including both the SQRT(-2) and the SQRT(2) extensions. However, not all can: if you extend the integers using SQRT(-5), you get 6 = 2*3 = (1 + SQRT(-5))(1 – SQRT(-5)). All of these four factors have no factors in the ring except themselves, 1, and their associates; it’s not like 12 = 2*6 = 4*3, in which case the underlying factorization is 2*2*3.

Without unique factorization, the entire argument that both factors of a cube must be cubes if they have no factors in common falls apart. This is because what this argument relies on is writing the cube as a product of its prime factors. In plain English, 216 = 2*2*2*3*3*3; taking the factors 8 and 27, we get 8 = 2*2*2 and 27 = 3*3*3, so 8 and 27 are cubes.

Obviously, this doesn’t work anymore when there is more than one factorization. In the SQRT(-5) extension, 36 is a square, but factoring it as 2*3*(1 + SQRT(-5))(1 – SQRT(-5)), we get that 2*(1 + SQRT(-5)) and 3*(1 – SQRT(-5)) have no common factors except units, but neither is square.

[…] A while ago, I wrote about unique factorization, and how it can sometimes fail. Today I want to introduce one concept that to some degree is intended to get around the failure of unique factorization in some environments: ideals. […]