It's easy, even for a pure mathematician, to sit through a
lecture on abstract algebra wondering why you should care. After
all, groups, rings, fields, and so on, seem to be fairly
arbitrary concepts, and the theorems you actually start proving in
the abstract are often, when applied to a particular case, much
simpler to prove by an alternative means. So I'm here to give a better
example of the power of (in this case) ring theory. You'll need to
know what a (commutative with identity) ring is, and to understand
the following statement: "If R is a Euclidean domain, then it is a
principal ideal domain (PID); and if R is a PID, then it is a unique
factorization domain (UFD).". Here goes.

We want to find all solutions in integers to the equation x^{2}+2=y^{3}. A quick bit of thought gives x=±5, y=3, but are there any others? The answer to this is in fact "no". Let's see why.

First, Z(i√2) is a Euclidean domain with the Euclidean valuation f(a+ib√2)=a^{2}+2b^{2} (this is just the square norm of the complex number a+ib√2, so this
is easy to check). Hence it's a UFD. Note that the only units are ±1,
so I'll ignore signs where relevant from now on. We can factor our
equation as y^{3}=(x+i√2)(x-i√2). If c divides both of the
factors on the RHS, it divides their difference 2i√2, so it must be
one of 1, 2, i√2, or 2i√2. Now it's easy to see that x must be odd
(just reduce the original equation mod 4). Thus c must be 1. That is,
the two factors are coprime, and since their product is a cube
(and the units are both cubes), each individual factor must also be a
cube. Putting x+i√2=(g+ih√2)^{3}, we obtain 1=3g^{2}h-2h^{3}, so h=±1. If h=-1, there's no integer g which works, so h=1 and g=±1. Thus x=g^{3}-6gh^{2}=±5, and thus y=3. Both these solutions work, and they are the only ones.

Update: Thanks, whoever softlinked "cock"..