MHB Equivalent Goldbach proof impossible question.

AI Thread Summary
The discussion centers on the challenge of proving that the number of unordered partitions of an even number 2n into two composite numbers exceeds that of an odd number 2n+1 into two composites for n greater than 1 and n not equal to a prime. Participants express skepticism about the feasibility of solving this problem, likening it to the unsolved Goldbach Conjecture. One contributor humorously suggests that the question is essentially a joke, indicating its complexity and the improbability of finding a solution. The thread highlights the difficulty of the mathematical challenge posed. Overall, the consensus is that the proof is unlikely to be achieved.
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Prove that the number of unordered partitions of an even number 2n into 2 composites is greater than the number of unordered partitions of an odd number 2n+1 into 2 composites for n>1 and n\ne p prime.
 
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Re: Equivalent Goldbach proof impossible queston.

Are you:

  • Posting this as a challenge (where you have solved it)? If so, I will move this thread to our challenges forum.
  • Asking for help with the question? If so, please post what you have done and where you are stuck.
 
Re: Equivalent Goldbach proof impossible queston.

MarkFL said:
Are you:

  • Posting this as a challenge (where you have solved it)? If so, I will move this thread to our challenges forum.
  • Asking for help with the question? If so, please post what you have done and where you are stuck.

No, this is equivalent to proving Goldbach Conjecture, no one is going to solve it. Just a joke. You can delete if you wish.
 

Thread 'Trigonometry problem of interest'

Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
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