4

For a given $n$, consider the assertion:

$\exists a \in \Bbb Z : a^n-(a-1)^n\ \text{is prime}\tag*{}$

How can one do one of the following:

  • Prove that the assertion is true for all integer $n > 1$
  • Prove that the assertion is false for all integer $n > 1$
  • Characterize the values of $n > 1$ for which the assertion is true

The assertion is clearly true (with $a=2$) for all values of $n$ that are Mersenne prime exponents. Also, we only need to consider positive values for $a$. If $a<0$ then $a^n-(a-1)^n$ is negative (and hence not prime) when $n$ is even and is equal to $(1-a)^n-((1-a)-1)^n$ when $n$ is odd.

I suspect the assertion is true for all $n > 1$, but I have no idea how to prove it.

EDIT (thanks to @Malcolm):

The assertion is false whenever $n$ is composite. This is because $a^{uv} - (a-1)^{uv}$ is divisible by $a^u - (a-1)^u$ and by $a^v - (a-1)^v$. (This can be shown by induction on $v$ for $a^u - (a-1)^u$ and vice versa.)

I'd still like to know if the assertion is true for $n$ prime and not a Mersenne prime exponent.

EDIT 2:

It turns out that this family of numbers has been studied before, although not extensively, as one form of generalized Mersenne numbers. (See, for instance, the 2003 conference paper Generalized Mersenne prime numbers: characterization and distributions by Vladimir Pletser.) It also turns out that there are indeed primes of the form $a^n - (a-1)^n$ for values of $n$ that are not Mersenne prime exponents. For instance,

$M_{11} = 2^{11} - 1 = 2047 = 23\cdot89\tag*{}$

while

$6^{11} - 5^{11} = 313968931\tag*{}$

is prime. Pletser established by brute force that there are primes of the form $a^n - (a-1)^n$ for all prime $n$ through 31. He and T. D. Noe have tabulated the smallest such prime for the first 80 primes. See OEIS A121620.

Ted Hopp
  • 585
  • To get you started, observe that for $n=2,3$ the value $a=2$ satisfies the condition. – pwerth Jan 16 '19 at 18:42
  • @pwerth - Right. However, I'm not particularly interesting in specific values of $n$. The assertion is obviously true (with $a=2$) for all exponents corresponding to Mersenne primes. I suspect that the assertion is true for all $n>1$, but I have no idea how to prove that. – Ted Hopp Jan 16 '19 at 18:52
  • I only wrote that comment because your question does not indicate that you had evaluated any base cases or even knew that it was true or false for a given $n$. I suggest rewording your question to include basically what you just wrote in that comment. – pwerth Jan 16 '19 at 18:54
  • @pwerth - Fair enough. I've updated the question. – Ted Hopp Jan 16 '19 at 19:04
  • 1
    The assertion fails for $n=4$ and for $n=6$ as $a^4 - (a-1)^4 = (2a-1)(2a^2-2a+1)$ and $a^6 - (a-1)^6 = (2a-1)(3a^2-3a+1)(a^2-a+1)$ – Malcolm Jan 16 '19 at 20:26
  • @Malcolm - Thanks for that. I wonder if it fails for all composite $n$. – Ted Hopp Jan 16 '19 at 20:34
  • @TedHopp I think it might. This may be related to factoring of cyclotomic polynomials, but it is a long time ago that I last looked at that. – Malcolm Jan 16 '19 at 21:01
  • 3
    @Malcolm - It turns out that it's easy to prove that $a^{uv} - (a-1)^{uv}$ has factors $a^u-(a-1)^u$ and $a^v - (a-1)^v$. – Ted Hopp Jan 16 '19 at 21:04
  • 1
    A step in the right direction would be to show $x^p-(x-1)^p$ is an irreducible polynomial for $p$ prime. – Cheerful Parsnip Jan 16 '19 at 21:22
  • If you can show a polynomial is irreducible, then under some other mild conditions, there is a conjecture that it achieves prime values infinitely many times. – Cheerful Parsnip Jan 17 '19 at 05:03
  • https://en.wikipedia.org/wiki/Bunyakovsky_conjecture – Cheerful Parsnip Jan 17 '19 at 05:48

0 Answers0