Diophantine equations and Hilbert's 10th Problem, how did MRDP do it? I'm having a bit of trouble understanding the Wiki explanation of MRDP's (Matiyasevich, Robinson, Davis, Putnam)'s Theorem, which explains that Hilbert's 10th problem is unsolvable.

The MRDP theorem states that a set of integers is Diophantine if and only if it is computably enumerable. A set of integers $S$ is recursively enumerable if and only if there is an algorithm that, when given an integer, halts if that integer is a member of $S$ and runs forever otherwise.

Source: http://en.wikipedia.org/wiki/Diophantine_set
What exactly is the set $S$ in the case of Hilbert's 10th Problem? Is it the set of solutions $n$ for any ${x_1}^n+{x_2}^n+...+{x_{k-1}}^n={x_k}^n$ Diophantine equation? 
Also are we assuming that the $x_1,...,x_k$, are unchanging? That is, we choose this list of constants $(x_1,x_2,...,x_k)$, and we look for all the $n$ that make the equation true?
MOST confusing to me: why does it matter if this set is c.e.? 
Sorry for so many questions, I've been sifting through Wiki pages and haven't been able to find a clear explanation for this.
 A: Matiyasevich showed that for any recursively enumerable set $S$ of natural numbers, there is a polynomial $P_S(y,x_1,x_2,\cdots,x_k)$ with integer coefficients such that for any non-negative integer $y$, we have 
$$y\in S \quad\text{if and only if}\quad \exists x_1 \exists x_2\cdots\exists x_k\left(P_S(y,x_1,x_2,\dots, x_k)=0\right)   .\tag{1}$$
Here the variables $x_i$ range over the non-negative integers, but one can let them range over the integers. 
The polynomial $P_S$ depends on $S$. However, one can show that there is a polynomial $P(e,y,x_1,\dots,x_k)$ such that for any recursively enumerable subset $S$ of the natural numbers, there is a natural number $e_S$ such that 
$$y\in S \quad\text{if and only if}\quad \exists x_1 \exists x_2\cdots\exists x_k\left(P(e_S, y,x_1,x_2,\dots, x_k)=0\right)   .\tag{2}$$
Let $y\in S$. Typically, the $(x_1,x_2,\dots,x_k)$ (there may be many) such that $P_S(y,x_1,\dots,x_k)=0$ will depend on $y$. And there are infinitely many $P_S$ that do the job of "representing" $S$ in the sense of Formula (1). 
The condition that $S$ be recursively e
numerable is necessary. This is because for any polynomial $P(y,x_1,x_2,\dots,x_k)$ with integer coefficients, the set 
$$\{y: \exists x_1\cdots \exists x_k(P(y,x_1,\dots,x_k)=0)\}$$
is recursively enumerable. 
Remarks: $1.$ Technically, the example $x_1^n+\cdots+x_{k-1}^n=x_k^n$ is not a Diophantine equation in the sense of the theorem, since a variable occurs as an exponent. It is an exponential Diophantine equation. The fact that every recursively enumerable predicate is exponential Diophantine was known before the work of Matiyasevich.
$2.$ It has been known since the time of Godel that there is a recursively enumerable predicate 9set) $S$ which is not recursive. Let $P_S(y,x_1,\dots,x_k)$ be a polynomial that represents this $S$ in the sense of Formula (1). Then there is no algorithm that will determine, given input $y$, whether the equation $P(y,x_1,\dots, x_k)=0$ has a solution in natural numbers. It follows that there is no algorithm for determining, given a Diophantine equation $Q(z_1,\dots,z_n)=0$, whether the equation has a solution in natural numbers. For if there were such an algorithm, it could be used in particular to determine, given $y$, whether $P_S(y,x_1,\dots,x_k)$ has a solution $(x_1,\dots,x_k)$.  
A: For a bit more, you might find some interest in this handout I did for a student reading group: http://www.logicmatters.net/resources/pdfs/MRDP.pdf
