What this is giving you is the series form of your expression in terms of p. It did that, rather than giving you an integer or rational, because you didn't tell it what p was. If you tell it that, it will compute the answer normally.

[QUOTE=CRGreathouse;230498]What this is giving you is the series form of your expression in terms of p. It did that, rather than giving you an integer or rational, because you didn't tell it what p was. If you tell it that, it will compute the answer normally.[/QUOTE]

I was just wanting an equation that maybe we could say simply it will be integer with this p but s is involved so I can't

Wrong thread
 

I just relized this was the wrong thread lol.

back again
 

I've been thinking about Goldbach's conjecture tonight and I can make 3 categories of number to prove:

1)2*prime (note: equals p+p)
2) x*2*prime
3) powers of 2 ( Note: merges with class 2 if (2^x)>4 )

So since we know 4 works it's #2 we have to test agreed ? I'll test these out to check I've got it correct and, Yes I realize #3 can is a mix between 1 and 2 and I realize #1 is just a special case of #2 with x=1 and prime=2.

[QUOTE=science_man_88;245779]I've been thinking about Goldbach's conjecture tonight and I can make 3 categories of number to prove:

1)2*prime (note: equals p+p)
2) x*2*prime
3) powers of 2 ( Note: merges with class 2 if (2^x)>4 )

so since we know 4 works it's #2 we have to test agreed ?[/QUOTE]

You're looking at numbers of the form 2n and saying that n is (1) prime, (2) an integer, or (3) a power of two. I agree that this is true for integers n (with some overlap between categories, of course) but I don't see how it gets you any closer to a solution.

[QUOTE=CRGreathouse;245781]You're looking at numbers of the form 2n and saying that n is (1) prime, (2) an integer, or (3) a power of two. I agree that this is true for integers n (with some overlap between categories, of course) but I don't see how it gets you any closer to a solution.[/QUOTE]

maybe it allows more formula to be exploited to a degree that we can find a formula for a solution for each x, for example for x = 1 p+p works. is there anything to be said about other x ? if so can we come up with general formula given an x ?

All cases are covered by your case #2, so there's no simplification to be had there.

[QUOTE=CRGreathouse;245792]All cases are covered by your case #2, so there's no simplification to be had there.[/QUOTE]

okay can we complicate it with a formula linking x and possible values of the primes involved knowing prime ?

another starting point I see is noting odd prime + 2 = odd number and odd + odd = even so we know that without being distinct all even numbers should be a sum of 4 not necessarily distinct primes, if and only if all odd + odd covers all even numbers. then it's limiting it to two from there. But I'm guessing this won't help either.

[QUOTE=science_man_88;245798]another starting point I see is noting odd prime + 2 = odd number and odd + odd = even so we know that without being distinct all even numbers should be a sum of 4 not necessarily distinct primes[/QUOTE]

That doesn't follow. Yes, the sum of two odd primes is even, but that doesn't mean that all even numbers are the sum of two odd primes.

Consider: the sum of two Mersenne primes is even, but not all numbers are the sum of two Mersenne primes.

[QUOTE=CRGreathouse;245814]That doesn't follow. Yes, the sum of two odd primes is even, but that doesn't mean that all even numbers are the sum of two odd primes.

Consider: the sum of two Mersenne primes is even, but not all numbers are the sum of two Mersenne primes.[/QUOTE]

I know, that's not what I was saying. I never said anything about just adding 2 odd primes. If you look I say 2 + odd prime = odd and odd + odd = even not odd prime + odd prime = even . if we prove odd prime +2 can cover all odds ( doesn't appear to) and odd+ odd = even covers all even then couldn't we deduce odd prime +odd prime = even ? I'll show you an example if you like.