<jpierre.pratali@w...> wrote:>

43

> 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 =

> + 53

or

> 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

> rational-gaussian).

(r)

>

> 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

> + 31(g) = 19(g) + 29(r)

(g)

> 50(h) = 3(g) + 47(g) = 7(g) + 43(g) = 13(r) + 37(r) = 19

> + 31(g)

(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

> + 61(r)

(r)

> 94(h) = 5(r) + 89(r) = 11(g) + 83(g) = 23(g) + 71(g) = 41

> + 53(r) = 47(g) + 47(g)

(g)

> 96(m) = 7(g) + 89(r) = 13(r) + 83(g) = 17(r) + 79(g) = 23

> + 73(r) = 37(r) + 59(g) = 43(g) + 53(r)

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

>

> Does a large table confirm (1) and (2) indefinitely ?

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"