This will be a post illustrating how unique factorization in the ring helps us solve the diophantine equation . We first start by proving

**Lemma:** The ring is a UFD.

**Proof.** Define . The identity

proves that

So is a multiplicative function. We claim that is a Euclidean norm on . Let with . We need to show that there exist elements such that where . Now , so we can write where . Let be the closest integers to and , respectively. Let . Since and , it follows that

Set . Then and

as desired.

Let’s turn our attention in solving the equation in integers and . Suppose satisfy . First, cannot be even because then which is impossible. So must be odd, and as a result is also odd. Now we have the factorization in . We first show that and are relatively prime. Suppose divides both and . Then divides their difference . If is not a unit, then would divide . This is because is prime: indeed, if , then from , we see that or ; so either or is a unit; i.e is irreducible and hence prime because we are in a UFD. But if divides , then divides and so divides . But this means , i.e. and . But then would be even, contradiction. We conclude that is a unit.

So and are relatively prime. So if is any prime that divides , then divides . So divides either or . Hence, must be of the form for some where is a unit. Since the only units in are (units correspond to ) which are already cubes, we can absorb them into term. So write for some . Then we get

As a result, and . We deduce that . This gives . Clearly doesn’t give any solution in integers; so and consequently, . So . Now gives . Therefore, are the only solutions in integers.