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Christine
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 prime powers   « on: Jul 18th, 2014, 2:15pm » Quote Modify

I found many examples of the sum of two primes equal to a prime power

p + q = r^n ............(p, q, r) are primes

Is the number of solutions finite or infinite?

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Grimbal
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 Re: prime powers   « Reply #1 on: Jul 18th, 2014, 2:49pm » Quote Modify

If q is 2, and n 1, it is the prime pair conjecture.
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rmsgrey
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 Re: prime powers   « Reply #2 on: Jul 21st, 2014, 4:33am » Quote Modify

At least one of p, q, r must be 2 for this to work at all.
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dudiobugtron
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 Re: prime powers   « Reply #3 on: Jul 21st, 2014, 5:43pm » Quote Modify

When r is 2, then the existence of solutions for every n follows from Goldbach's Conjecture.
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towr
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 Re: prime powers   « Reply #4 on: Jul 21st, 2014, 10:22pm » Quote Modify

Now we just need to prove the Goldbach conjecture and we're done!
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0.999...
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 Re: prime powers   « Reply #5 on: Jul 22nd, 2014, 2:43am » Quote Modify

There's a chance Zhang's recent result on the distribution of the primes numbers could help, though the result and this riddle aren't exactly the same.
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Christine
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 Re: prime powers   « Reply #6 on: Jul 23rd, 2014, 10:20am » Quote Modify

on Jul 21st, 2014, 4:33am, rmsgrey wrote:
 At least one of p, q, r must be 2 for this to work at all.

p + q = r^n
(p, q, r) are primes

if p > 2, q > 2, (p,q) are odd primes,  then r = 2

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towr
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 Re: prime powers   « Reply #7 on: Jul 23rd, 2014, 12:14pm » Quote Modify

Because 2 is the only even prime.
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rmsgrey
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 Re: prime powers   « Reply #8 on: Jul 24th, 2014, 5:17am » Quote Modify

on Jul 23rd, 2014, 10:20am, Christine wrote:
 p + q = r^n (p, q, r) are primes   if p > 2, q > 2, (p,q) are odd primes,  then r = 2   could you please explain why?

towr's already hit the key point, but:

If p, q, and r are all odd, then r^n is also odd, but the sum of two odd numbers is an even number. Contradiction.

Therefore at least one of p,q,r must be even (in fact, either one or all three of them must be) but they're also prime, so at least one must be an even prime, and, since all even numbers are divisible by two, the only one that's prime is two itself.

There's a solution where p, q, r, n are all 2, and solutions where exactly one of p, q, r is 2, and no other solutions.
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