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        riddles >> medium >> prime powers
        
(Message started by: Christine on Jul 18^th, 2014, 2:15pm)

Title: prime powers
Post by Christine on Jul 18^th, 2014, 2:15pm

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?

Title: Re: prime powers
Post by Grimbal on Jul 18^th, 2014, 2:49pm

If q is 2, and n 1, it is the prime pair conjecture.

Title: Re: prime powers
Post by rmsgrey on Jul 21^st, 2014, 4:33am

At least one of p, q, r must be 2 for this to work at all.

Title: Re: prime powers
Post by dudiobugtron on Jul 21^st, 2014, 5:43pm

When r is 2, then the existence of solutions for every n follows from Goldbach's Conjecture.

Title: Re: prime powers
Post by towr on Jul 21^st, 2014, 10:22pm

Now we just need to prove the Goldbach conjecture and we're done! :P

Title: Re: prime powers
Post by 0.999... on Jul 22^nd, 2014, 2:43am

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.

Title: Re: prime powers
Post by Christine on Jul 23^rd, 2014, 10:20am

on 07/21/14 at 04:33:24, 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

could you please explain why?

Title: Re: prime powers
Post by towr on Jul 23^rd, 2014, 12:14pm

Because 2 is the only even prime.

Title: Re: prime powers
Post by rmsgrey on Jul 24^th, 2014, 5:17am

on 07/23/14 at 10:20:35, 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.