## Re: Goldbach : Two new conjectures inside the first ?

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• ... 43 ... or ... (r) ... (g) ... (g) ... (r) ... (g) ... The answer is that we do not need any experimental checking. Every even number is of the form 4k or
Message 1 of 2 , Feb 23, 2005
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<jpierre.pratali@w...> wrote:
>
> According to the Goldbach conjecture every even integer >2 is, by
> one or several manners, the sum of two primes.
> I suppose it exists a large table of such decompositions for a wide
> range of even numbers. Personally I could not find where.
> So I built, "by hand", a small table for even numbers :
> ...
> 14 = 3 + 11 = 7 + 7
> 16 = 3 + 13 = 5 + 11
> 18 = 5 + 13 = 7 + 11
> ...
> 46 = 3 + 43 = 5 + 41 = 17 + 29 = 23 +23
> 48 = 5 + 43 = 7 + 41 = 11 + 37 = 17 + 31 = 19 + 29
> 50 = 3 + 47 = 7 + 43 = 13 + 37 = 19 + 31
> ...
> 90 = 3 + 87 = 7 + 83 = 11 + 79 = 17 + 73 = 19 + 71 =
> 23 + 67 = 29 + 61 = 31 + 59 = 37 + 53 = 43 + 47
> 92 = 3 + 89 = 13 + 79 = 19 + 73 = 31 + 61
> 94 = 5 + 89 = 11 + 83 = 23 + 71 = 41 + 53 = 47 + 47
> 96 = 7 + 89 = 13 + 83 = 17 + 79 = 23 + 73 = 37 + 59 =
43
> + 53
> 98 = 19 + 79 = 31 + 67 = 37 + 61
>
> Now let us examine each set of decomposition for each even.
> We remember that a prime is either a "gaussian" one (-1 mod 4) or
> a "not gaussian" one (1 mod 4) (they also call it "rational"; it is
> also the sum of 2 squares )
>
> So let us look at the pairs of primes for each even : We can see
> that
>
> 1. the corresponding pairs are either all "homogeneous" (gaussian-
> gaussian or rational-rational) or all "mixed" (gaussian-rational
or
> rational-gaussian).
>
> Next, if we consider the set of all the even numbers >4, we can see
> that
>
> 2. a "mixed" (m) always succeeds to a "homogeneous" (h) and
> inversely.
>
> For example (with (g) for "gaussian", (r) for "rational", (h) for
> homogeneous" and (m) for "mixed" :
>
> 46(h) = 3(g) + 43(g) = 5(r) + 41(r) = 17(r) + 29(r) = 23g)
> + 23(g)
> 48(m) = 5(r) + 43(g) = 7(g) + 41(r) = 11(g) + 37(r) = 17
(r)
> + 31(g) = 19(g) + 29(r)
> 50(h) = 3(g) + 47(g) = 7(g) + 43(g) = 13(r) + 37(r) = 19
(g)
> + 31(g)
> 52(m) = 5(r) + 47(g) = 11(g) + 41(r) = 23(g) + 29(r)
> ...
> 90(h) = 3(g) + 87(g) = 7(g) + 83(g) = 11(g) + 79(g) = 17
> (r) + 73(r) = 19(g) + 71(g) = 23(g) + 67(g) = 29(r) + 61(r)
> = 31(g) + 59(g) = 37(r) + 53(r) = 43(g) + 47(g)
> 92(m) = 3(g) + 89(r) = 13(r) + 79(g) = 19(g) + 73(r) = 31
(g)
> + 61(r)
> 94(h) = 5(r) + 89(r) = 11(g) + 83(g) = 23(g) + 71(g) = 41
(r)
> + 53(r) = 47(g) + 47(g)
> 96(m) = 7(g) + 89(r) = 13(r) + 83(g) = 17(r) + 79(g) = 23
(g)
> + 73(r) = 37(r) + 59(g) = 43(g) + 53(r)
>
> Does a large table confirm (1) and (2) indefinitely ?

The answer is that we do not need any experimental checking.

Every even number is of the form 4k or 4k+2
If the even number is 4k=4(k1+k2+1)=(4k1+1)+(4k2+3)
If the even number is 4k+2=4(k3+k4)+2=(4k3+1)+(4k4+1)
4k+2=4(k5+k6+1)+2=(4k5+3)+(4k6+3)

So even 4k+2 are always "homogeneous" and even 4k are always "mixed"
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