Math Is Fun Forum

I am working on a simpler theory;

e=any even no.
e=(c-a)+(d-b)

1) primes <e excluding 2
2) 2
c and d factorable by 1) not factorable by 2)
a and b factorable by 2) not factorable by 1)
Therefore (c-a) and (d-b) are prime.

I am not however sure whether I need to prove that e will = a certain size....?

More like this;
x is not factorable by 2 or 5.
x = ab

x ends in 1; a and b end in (1,1)(3,7) or (9,9).
x ends in 3; a and b end in (1,3) or (7,9).
x ends in 7; a and b end in (1,7) or (3,9).
x ends in 9; a and b end in (1,9)(3,3) or (7,7).

i.e.  3x9=27.......ends in 7.
by knowing what x ends in we can determine what a and b might end in.

You would have thought knowing what a and b end in you could determine that a number is composite and therefore not prime, turns out it doesn't seem to work that way.

Yes. But should be able to work out a rule to find out the number of primes up to a limit.

Say x=29,

sqrt(29)=5 rd. dwn. to nearest prime,

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 Delete All Even No.'s,
1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 Times remaining by 3= 3,9,15,21,27,33 etc.......Remove these:
1 5 7 11 13 17 19 23 25 29     Remaining No.'s Timesed by 5 = 5,25,35 etc. Remove theses again = 1 7 11 13 17 19 23 29.

Take all primes < or = to sqrt.(x).

I have edited formula. Thanks.

No, what I am saying is only Primes and 1 can not have (x - y)>1, when

If when (a) is prime, (b) always isn't, that would mean that (b) would have a prime within 12 numbers of b and not more. If it did have a bigger gap than 12 then there would not be a prime occurring every 13 numbers.
If primes are occurring within 12 numbers of b and a is occurring 13 times in 169, then non - primes (b) = (h,i,j,k,l,m,n,p,r,t,u,v,w,) fill in each gap with a prime and you get 14 primes i.e.don't forget h minus up to 12. Therefore one gap must =0 for there to be 13 primes and not 14. I.e. there must be at least one occurrence of (b) being prime when (a) is.

The Algorithm will always work because 13 primes with the biggest gap between them being 13-1=12 means that 13 squared (169) is the biggest range you would need to work with to find one occurrence of a and b.

I think I might have made a mistake which makes things simpler:
For No.'s<(z)squared the greatest gap between two primes will be (z - 1).
This is because the greatest prime we are concerned with is not z, it is the prime below z. Therefore you do not need to be concerned with remainders that are as big as z.

e.g. for primes <49 need only have remainders; when you try to factor primes 2,3 and 5.
You need not be concerned with a remainder >6 because; that would incorporate 7. We are not concerned with the prime 7 having a remainder at all.

Two numbers that add up to e can be prime because there is 1/z occurences of primes in (z)squared therefore the occurences of two random numbers being prime is 1 in (z)squared.

(1) All I mean is; say e = a + b and there was only 1 prime that could = a and one prime that could = b. Then there would be only one occurrence of this happening in e and that's all we are looking for...at least one.

(2) Secondly (1/sqrt(e))*(1/sqrt(e)) would equal this.
If you round down......then you will equal a No. greater than sqrt(e)...........which means you can then say that the percentage must equal this. I can explain it as follows; if there were one prime occurring every 5 no.'s in the whole of 100. Then when I half 100..I get 50 and if there's one occurring every 5 No.'s on one side and one prime occurring every 5 No.'s on the other. Then +/- c = a or b, and a and b must occur because the probability of c causing a and b to equal primes is the above probability. There's a physics formula where if the No. of trees in a forest reaches a certain percentage, then if one of the trees catches fire they all do.

(3) Thirdly the last question is really complicated to explain but is all true:

To get a No. not factorable by a bunch of primes, you have to get those numbers multiplied together. Then minus....or add, a no. NOT factorable by them. Once you've grasped that, it's simple, to get these formula's such as 2m +/- 1 or 6m+/- 1 or 5 you just repeat the formula (x) no. of times and then remove all no.'s factorable by x....which will be exactly the PREVIOUS set multiplied by x. All primes will be found in < ((the next prime above (x))squared) otherwise they are at risk of being factorable by primes>x.

2m+/-1
6m+/-1 or 5
30m+/- 1 or 7 or 11 or 13 or 17 or 19 or 23 or 29
210+/- 1 or 11......all no.'s not factorable by 2,3,5 or 7....(therefore prime if<11 squared)

(4) Also the gap that I talked about alters the percentage possibly, but everywhere around that gap will have the correct percentage. e +/- c = a or b, how big does c have to be? sqrt(e) rd. dwn. I think because then you will have enough occurences for probability to take effect. This is just right because the big gap will occur exactly at (x) squared. So maybe round e down to the nearest PRIME and it's true.

Thanks.