Problems & Puzzles: Conjectures

Conjecture 47. P=2P1 + 3P2

Vassilis Papadimitriou sent the following Conjecture:

Every prime number p>3 can be written as p = 2*p1 + 3*p2, where p1, p2 are primes or 1?

Question: Prove it or show it false.


Contributions came from Farideh Firoozbakht & Luis Rodriguez.


Farideh wrote:

I think a better form of this conjecture is:
            " Every prime number p=>19 can be written as p=2*p1 + 3*p2 
              where p1 &  p2 are odd primes. "
Note that 19 is the smallest number of the form 2*p1 + 3*p2 where p1 & p2 are
distinct odd primes.
It seems that like the Goldbakh's conjecture,  the proof  is difficult or impossible
at this age. 


Rodriguez wrote: Gauss said: "I can frame infinitely many propositions of that type, that never will be contradicted" (*)

This one is much simpler: "Every p=p2-p1 +1"

Or this: "p = p1 + p2 + p3 ; p>7", proved by Vinogradov.

Or this :"Every p = p(n+1)- p(n) + 1"

Or this :"For every p there is an n >=0 that makes 6(p - n^2) + 1 a prime.

(*) I confess that Fermat's Theorem as an isolated proposition has very little interest for me, because I could easily lay down a multitude of such propositions, which one could neither prove nor dispose of.[A reply to Olbers' attempt in 1816 to entice him to work on Fermat's Theorem.].Quoted in J R Newman, The World of Mathematics (New York 1956).



Records   |  Conjectures  |  Problems  |  Puzzles