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.