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Does anybody know if this is true?

I can't find references about it, also I can't prove to be true (or false).

I think computing $x$ is a brute force task.

Thanks (and sorry if I lost some basic concept).

Ken
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Miguel Velilla
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  • Do you mean positive integers? $q=-1, n=1$ clearly has no solution – nosyarg Jun 10 '15 at 02:28
  • This is almost certainly unknown, unless there is a trivial class of counterexamples which are not immediately obvious to me. In that case, determining for which $x$ this is possible would almost certainly be unknown. We cannot use density estimates, as primes are too sparse. We do not understand densities of primes along any nonlinear polynomial, for that matter. It's not even clear to me that it should or shouldn't be true. – davidlowryduda Jun 10 '15 at 02:30
  • @mixedmath I would think that one could generate a counterexample using difference of squares, cubes etc. but I am at a loss at the moment – nosyarg Jun 10 '15 at 02:34

2 Answers2

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Definitely not for all pairs $(q,n)$.

For instance $x^q-1$ is a multiple of $x-1$ and so if $x^q-1$ is prime then $x-1=1$ and so $x=2$. But $2^q-1$ is prime only if $q$ is prime. So, for all $q$ composite, $x^q-1$ is never a prime.

Finding the $q$ for which $2^q-1$ is prime is major undertaking and the main source of large primes. See Mersenne prime.

lhf
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  • Ok sorry my lack of precision. There are lot of counter examples when n is a power with some common factors on q (to say for example q = 3 and n = 8 = 2^3). These are "trivial cases" which clearly lead to non-prime for all x. – Miguel Velilla Jun 10 '15 at 02:46
  • I think this statement could be reinforced by adding some constrains, but my intention is only to pass the central idea. Some times it is hard to find the prime which solves it, for instance to solve n=15 and q=6 the lower prime I found was 232218265089212401 and x=784 – Miguel Velilla Jun 10 '15 at 02:58
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It is false for $q=3$ and $n=12^3$, since $x^3-12^3 = (x-12)(x^2+12x+144)$, which is negative and so not prime for $x<12$; equals $0$, not prime, for $x=12$; equals $469 = 7 * 67$ for $x=13$, and finally is divisible by $x-12$ for $x>13$.

jbuddenh
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