Can $x^n-(x-1)^n$ be Prime if $n$ is Not Prime? I'm hoping someone can provide an answer or a link to a proof regarding this question.
Edit:
The question has been put on hold because I did not expound on why the answer to this was of interest to me or the community, so, even though I've received my answer, I will elaborate.
In my spare time I've been attempting to develop novel ways of generating odd compound numbers as well as primes. A pattern emerged as I iterated across $\mathbb{n \in N}$. Prime numbers were only turning up where $\mathbb{n \in P}$.

Let's see.....
$a^n - b^n = (a-b)(a^{n-1} + a^{n-2}b + .... + b^{n-1})$ so
$x^n - (x-1)^n = (x- (x-1))(x^{n-1} + x^{n-2}(x-1) + ... +(x-1)^n)$
$= 1\times(x^{n-1} + x^{n-2}(x-1) + ... +(x-1)^n)$.
Heh, heh.  That was cute.  I see what you did there.
...
But okay... this also means that if $n$ is not prime then 
$n = jk$ for some integers $j,k$. and 
$a^n - b^n = (a^j)^k - (b^j)^k = (a^j - b^j)(a^{(k-1)j} + a^{(k-1)j}b^j + ... + b^{(k-1)j})$.
Now $a^j - b^j$ cant be equal to $1$ because if $a \ge b+1$ the $a^j \ge (b+1)^j = b^j + jb^{j-1} + ..... + jb + 1 > b^j+1$.
(Oh, this assumes $a > b > 0$ which we might as well assume.  If $b > a$ we get negative results and we just need to switch the $a,b$ to get equivalent results.  And if $b = 0$ or $a = b$ the results are too trivial to bother with.)
And so neither $(a^j - b^j)$ nor $(a^{(k-1)j} + a^{(k-1)j}b^j + ... + b^{(k-1)j}$ are equal to $1$ so this isn't prime. 
Doesn't matter of $a = x$ or $b = x-1$ the result is the same:
