Problems & Puzzles: Puzzles

Puzzle 154.  Extension and Variation of the GC

Rober Burrage, poses in one of his Web pages an "Extensión of the Goldbach's Conjecture" (EGC). This EGC proposes that:  

"If N = A.B, then N can be expressed as a sum of A primes."

He also wonders if this his conjecture can be proved true supposing true the Goldbach Conjecture (GC).

It is a kind of easy to prove that the EGC follows directly from the GC, but I will let the readers to find that proof his own way. I will only provide - as indirect evidence that such proof exists - a code in Ubasic of mine to produce such representations for A & B >3.

10 input A,B:'goldbext.ub
20 P=2: if odd(B)=1 then P=3
30 Z=A*B-P*(A-2):Q=1
40 Q=nxtprm(Q):if (Z-Q)<>prmdiv(Z-Q) then goto 40
50 print A;"*";B;"=";
60 for I=1 to A-2:print P;"+";:next I
70 print Q;"+";(Z-Q):end

Some examples are:

5 * 12 = 2 + 2 + 2 + 7 + 47
4 * 15 = 3 + 3 + 7 + 47
12 * 9 = 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 5 + 73

Now I propose a variation of the Burrage's EGC that I will name VEGC:

"If N = A.B, then N can be expressed as a sum of A distinct primes under the condition X"

Questions:

a)    Does the VEGC follow also from the GC?

b)    What is that X condition?

c) Can you produce a code to make such expansion?

Solution

Luis Rodríguez has proved his own way the Burrage's conjecture (EGC):

Let N=A.B

a) Let N is even. Then N = (N -2(A-2)) +2(A-2). Then (N -2(A-2)) has the same parity than N, that is to say even. By the GC (N -2(A-2)) = p1 +p2 if (N -2(A-2))>3. Then N = p1+p2+2(A-2) if (N -2(A-2))>3 

b) N is odd. Then N = (N - 2(A-3) - 3) + 2(A-3) + 3. Then (N - 2(A-3) - 3) has opposite parity than N, that is to say even. By the GC (N - 2(A-3) - 3) = p1 + p2 if (N - 2(A-3) - 3)>3. Then N = p1+p2+2(A-3)+3.

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As you can see this correct proof provides another solution than the given by my code written above, specifically to the case where N is odd. In the Rodríguez solution for N=5x7 then A=5 and N=(35-2(5-3)-3)+2(5-3)+3=(28)+(2+2+3)=(11+17)+(2+2+3)

While using my code the solution is:

N=Z+3(5-2)=Z+(3+3+3)=26+(3+3+3)=(3+23)+(3+3+3)

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