Problems & Puzzles: Puzzles

Conjecture 88. Special Goldbach Partitions Ivan Ianakiev sent the following nice conjecture "Let p denote any prime in any decomposition of 2n. 4 and 6 are the only numbers n such that 2n + p is prime for every p". See OEIS A045917. Known examples:   8=3+5; then 8+3 & 8+5 are primes. 12=5+7; then 12+5 &12+7 are primes.   Q. Can you find another example or prove the conjecture?

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On May 27, 2024, John W. Nicholson wrote:

The question is equivalent to: 

 

For which n is  2n + p =/= 0 mod 5 for all p as defined, where =/= is not congruent. 
 

One may notice the double negative going on here caused by the 0 mod 5. 
 

For all p > 6 are of the form {1, 3, 7, or 9} mod 10. 

So, 1+1 = 2, 1+3 =4, 1+7 = 8 == 3 mod 5, 1 + 9 = 10 = 0 mod 5. 

For 3+1 = 4, 3+3 =6 == 1 mod 5, 3+7 = 10 == 0 mod 5, 3 + 9 = 12 = 2 mod 5. 

And for 7 + 1 = 8 == 3 mod 5, 7 + 3 = 10 == 0 mod 5, 7 + 7 = 14 == 4 mod 5, 7 + 9 = 16 == 1 mod 5. 

Lastly and, for 9 + 1 = 10 == 0 mod 5, 9 + 3 = 12 == 2 mod 5, 9 + 7 = 16 == 1 mod 5, 9 + 9 = 18 == 3 mod 5.Since all RC have at least one 0 mod 5  for all p the conjecture holds.

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On June 9, 2024, Ivan Ianakev wrote:

While I' m grateful for John W. Nicholson' s efforts to prove the conjecture, I do not think that my original definition is "equivalent to: For which n is 2*n + p =/= 0 mod 5 for all p as defined, where =/= is not congruent".

 

Let us take for example n = 25. The Goldbach parts of 2*25 = 50 form A = {3, 47, 7, 43, 13, 37, 19, 31}
and their sums with 50 form B = {53, 97, 57, 93, 63, 87, 69, 81}. No element of B is congruent to 0 mod 5, but that does not mean they are all prime (in fact, only 53 and 97 are prime). That is why n =
25 is not a solution. This, I think, is enough to invalidate John' s argument.

 

The following Mathematica code, based on John' s argument, produces many false solutions such as 2, 3, 8, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, ...

f[n_] := PrimePi[n/2];
g[n_] := Module[{lst = {}}, Do[If[PrimeQ[n - Prime[i]], AppendTo[lst, {Prime[i], n - Prime[i]}]], {i, 1, f[n]}]; lst];
Select[Range[2, 100], FreeQ[Mod[2*# + Flatten[g[2*#]], 5], 0] &]

 

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