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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

Jose Garcia
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    What? Which numbers are supposed to be fixed, and which are to be found? – Chris Eagle Mar 01 '13 at 21:02
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    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 $\mathbb P + \mathbb P$. –  Mar 01 '13 at 21:04
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    I think he means for each $n$, we can find $p,q,r$. – Yimin Mar 01 '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 01 '13 at 21:08
  • @ThomasAndrews, why what? by the way I didn't make any jokes - just said what I think about the problem. –  Mar 01 '13 at 21:10
  • I think Yimin had it right based on his examples, for each $n$, find a $p, q$ and $r$ prime. – muzzlator Mar 01 '13 at 21:12
  • Actually, you don't say, "I think ...." you say "no method on earth can prove it." That's what I mean by snark - it's not a rigorous assertion, but a statement of belief asserted as fact in a flip way. Hence the word "snark" that I chose. @user58512 My point is that you should be more precise when you are dealing with either beginners or people who are not adept in the language. – Thomas Andrews Mar 01 '13 at 21:12
  • If $n=1$,then one of $p,q$ should be 2, then it is the twin primes. if $n\ge 1$, we shall expect $p=2$ or $q=2$. Then the problem turns out to be something like twin prime problem. – Yimin Mar 01 '13 at 21:13
  • I don't think there is a general solution for all n, and even the n=1 statement would prove the twin prime problem. As noone can expect to get a full answer here, i flagged the question as their doesn't exist an answer – Dominic Michaelis Mar 01 '13 at 21:19
  • One of $p,q$ is $2$ no matter what $n$ is, if $n>0$. @Yimin – Thomas Andrews Mar 01 '13 at 21:22
  • n=0 has some easy solutions too though, n<0 has some issues as I don't think that is true in those cases as p then becomes a fraction. – JB King Mar 01 '13 at 21:52
  • Jose, you have asked over 150 questions here --- surely you know by now how to write one with enough detail so that people don't have to guess what you mean. – Gerry Myerson Mar 02 '13 at 06:38

1 Answers1

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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.

Yimin
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