Solutions in complex number field, instead of $\mathbb{N}$, to Fermat's Last Theorem As far as I know, we are searching solutions in set of positive integers for $x^n + y^n = z^n$  for $n > 2$. There are many proofs are stated that Fermat is true and there is no solutions for this equations. Now my question is, can we have solutions in complex numbers field? If yes, how to find them. 
Thanks in advance.

In the complex numbers, you can pick literally any $x$, $y$, and $n$ you want (with $x$ and $y$ complex).  Then compute $x^n + y^n$.  In the complex numbers this has $n$ different $n$th roots, unless it is zero, in which case it has only one $n$th root.  So pick any of these $n$ roots (or $0$ in the second case) for $z$, and you have a solution.
In the real numbers, the same method actually turns out to work.  As before, pick your favorite $x,y,n$ ($x,y$ real numbers now) and compute $x^n + y^n$.  If $n$ is odd, then every real number has a real $n$th root (for example every real number has a cube root) so we can pick this for $z$ and we're done.  If $n$ is even, we need the expression to be nonnegative in order to have a real $n$th root.  But $x^n$ and $y^n$ will both be $\ge 0$, so $x^n + y^n$ is nonnegative, so we can likewise pick $z$ to be the $n$th root and we run into no problems.  In fact in the even case there is a positive and a negative $n$th root, hence 2 options for $z$ given $x,y,$ and $n$.
