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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 $\mathbb{n \in P}$.

Jaarbahd
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3 Answers3

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The answer is no. Note that if $n=kl$ is the product of smaller numbers, then $$x^n-(x-1)^n \mbox{ is divisible by } x^k-(x-1)^k$$

More generally, $$a^n-b^n=(a^k)^l-(b^k)^l=(a^k-b^k)(\mbox{something})$$

cannot be prime unless $a^k-b^k=1$ or $a^n-b^n=a^k-b^k$.

J. W. Tanner
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N. S.
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Hint: $x^n - y^n$ is divisible by $x^r - y^r$ if $r$ divides $n$.

Robert Israel
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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:

fleablood
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