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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

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4  
What? Which numbers are supposed to be fixed, and which are to be found? –  Chris Eagle Mar 1 '13 at 21:02
2  
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
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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

1 Answer 1

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.

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