Diophantine equations solved using algebraic numbers? On Mathworld it says some diophantine equations can be solved using algebraic numbers. I know one example which is factor $x^2 + y^2$ in $\mathbb{Z}[\sqrt{-1}]$ to find the Pythagorean triples.
I would be very interested in finding some examples of harder equations (not quadratic) which are also solved easily using algebraic numbers. Thank you!

There are many examples, some of which I see you have already gotten :-)
Here is one example I had come across recently:

Show that the following diophantine equation,
  for integral $\displaystyle a,b$, has no non-trivial solutions.
$$3b^4 + 3b^2 + 1 = a^2$$

I came across this while trying to find an "elementary" solution to the diophantine equation
$$y^2 = x^3 - 1$$
which is easily solved using $\displaystyle \mathbb{Z}[i]$, see this here: Integral solutions to $y^{2}=x^{3}-1$
Of course, there might be an easy elementary proof for $\displaystyle 3b^4 + 3b^2 + 1 = a^2$, but, in fact this is closely related to the Pell's equation $\displaystyle 3x^2 + 1 = y^2$ and amounts to showing that the non-trivial odd values of $\displaystyle x$ are such that $\displaystyle \frac{x-1}{2}$ cannot be a perfect square. This itself can be cast into a problem of the terms of a linear recurrence never being perfect squares. So I do expect this problem to give some resistance to an elementary solution.
To solve the problem:
Mutiply by $\displaystyle 9b^2$
$$3b^2(9b^4 + 9b^2 + 3) = (3ab)^2$$ 
put $\displaystyle x = 3b^2 + 1$ and $\displaystyle y = 3ab$
$$(x-1)(x^2 +x + 1) = y^2$$
$$x^3 - 1 = y^2$$
which we can show has no non-trivial solutions easily, by using $\displaystyle \mathbb{Z}[i]$.
Note: If one can prove non-existence of $\displaystyle 3b^4 + 3b^2 + 1 = a^2$ by other means, we will have found a different proof of non-existence of $\displaystyle x^3 - 1= y^2$, because the existence of non-trivial solutions to $\displaystyle x^3 - 1= y^2$ implies the existence of non-trivial solutions to $\displaystyle 3b^4 + 3b^2 + 1 = a^2$. (That is how I came across this).
I will leave that to you.
