Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

let be $ p,q,r $ prime numbers AND 'n' an integer

is then true that we can always look for p,q,r and an integer n so

$$ p^{n}+q=r $$

  • $ 5+2=7$
  • $ 2^{3}+3=11 $
  • $ 3^{4}+2=83 $

abnd so on

share|cite|improve this question
What? Which numbers are supposed to be fixed, and which are to be found? – Chris Eagle Mar 1 '13 at 21:02
if you mean infinitely many solutions of the diophantine equation, then it's (1) obviously true (2) no method on earth can prove it. You could restate it: do infinitely many prime powers occur in the difference set $\mathbb P - \mathbb P$. Probably every number occurs. Goldbach asks about $\mathbb P + \mathbb P$. – user58512 Mar 1 '13 at 21:04
I think he means for each $n$, we can find $p,q,r$. – Yimin Mar 1 '13 at 21:08
@user58512 You should usually give some hint why, rather than what appears to be snark. From the nature of the question, the poster is either a beginner (and hence won't get subtle math jokes,) or not a native English speaker (with the same result.) – Thomas Andrews Mar 1 '13 at 21:08
@ThomasAndrews, why what? by the way I didn't make any jokes - just said what I think about the problem. – user58512 Mar 1 '13 at 21:10

If $n=1$, it is twin prime. Twin prime

If $n\ge 2$. we can see if $p=2$, then the problem is

$2^n+q = r$, actually it is Polignac's conjecture.Polignac

if $q=2$, then the problem is

$p^n+2 = r$, it is something like twin prime.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.