On Thursday, September 4, 2014 3:31:23 PM UTC-5, Archimedes Plutonium wrote: > Alright, let me begin making lists: > >
Of course we have 1+1=2
> > 1+4 =5 and we have 4-1=3 > > > > 4+9=13 and we have 9-4=5 > > > > 25+4=29 and we have 16-9=7 > > > > 25+16=41 and we have 36-25=11 > > > > 49+4=53 and we have 49-36=13 > > > > So it looks like we get about as many additions to make primes as subtraction. > > > > Now I wonder if the Pythagorean theorem has some role in this conjecture because as we notice the square roots of perfect squares: > > > > 5+2=7 from sqrt25+sqrt4 > > > > A proof of the Conjecture appears to be independent of the proof of Goldbach. >
Independent but using a key aspect of the AP-Postulate that between successive perfect-squares lies at least two primes.
I am confident the conjecture is true, which leaves for a intriguing proof in that as the primes get larger the spacing between primes increases and the spacing between perfect squares increases. For example at 3 to 5 is a spacing of 2 units but at 23 to 29 is a spacing of 6 units. And for perfect squares from 4 to 9 is a spacing of 5 but at 81 to 100 is a spacing of 19.
The baffling part of the conjecture and its proof when forthcoming is how and why perfect-squares are related to primes. It is not at all obvious that perfect squares are aligned in some orderly fashion with primes.
Now there is a Trivial Conjecture of primes and perfect-squares if we admit 0=0^2 as a perfect square, so that if you listed all the perfect squares 1, 4, 9, 16, 25, etc etc and then you have a formula of Prime = sqrt(perfect-square) +0 and deleting all nonprimes.
My conjecture is more substantial in that I have Prime = Perfect-square_1 + or - Perfect-square_2
The proof is not going to be easy.
This conjecture seems to be an original conjecture for I see nothing remotely similar in the literature.