20200823, 10:50  #1 
Aug 2020
Brasil
100000_{2} Posts 
Where is the weakness of this reasoning?
Let’s be any two positive and/or negative Odd Primes P1 and P2.
Because both are Odd Primes, there will always exist an Integer reference r equidistant from them such as: P1=r+d P2=rd being d the equal distance from the reference r to the both Odd Primes. For P1=P2, d=0. Then, P1+P2=2r r=(P1+P2)/2. And, P1P2=2d d=(P1P2)/2. If r=Odd, then d=Even and viceversa. OK. Now, the Semiprime formed with the two Primes above is P1P2=(r+d)(rd) So, P1P2=r^2d^2. Consequently, if any Odd number can be expressed as a difference of two Squares, and if any Odd difference between two Squares can be expressed as an Odd Semiprime number, then there will always exist 2 Primes P1 and P2 such as P1+P2=2r for any r. For P1=P2, the Semiprime is a Prime squared. And absolute value of d do not need necessarily be less than r. 
20200823, 12:42  #2 
"Viliam Furík"
Jul 2018
Martin, Slovakia
2·373 Posts 
I have noticed, that if you set the r = 2, you can't find a pair of primes that would satisfy, because at the beginning you said the primes have to be odd. But I think that's OK, you are only one case short to infinity
But I think the weakness is that you assume that EVERY number 2d can be expressed as a difference of two primes. As far as Google tells me, it has not been proven yet. Also, I am not sure about this part (...if any Odd difference between two Squares can be expressed as an Odd Semiprime number,...). If you manage to prove both my doubts, I am fairly certain you prove the Goldbach conjecture. 
20200823, 12:50  #3 
"Oliver"
Sep 2017
Porta Westfalica, DE
2^{3}·3·5·7 Posts 
That's only when you only allow positive primes. Otherwise, let P1=3, P2=7, then P1+P2=4=2*2=2r with your r=2.

20200824, 09:24  #4 
Romulan Interpreter
"name field"
Jun 2011
Thailand
5^{3}·79 Posts 
There is no weakness of the reasoning. The reasoning is sound.
What does it prove? (if you answer "Goldbach" then you still have to show us how you express 105 as a semiprime) (edit: ignore the other two guys above ) Last fiddled with by LaurV on 20200824 at 09:34 
20200824, 09:40  #5 
Romulan Interpreter
"name field"
Jun 2011
Thailand
5^{3}×79 Posts 
He never said that, which would be equivalent with G. He said difference of squares, which is true, every odd number is a difference of squares, because all squares can be obtained as partial sums of the 1+3+5+7+...., but the second part, about every odd being a semiprime... well.... I still can't write 105 as a semiprime, with all my efforts... maybe he teaches us how to do it.

20200824, 12:11  #6  
Feb 2017
Nowhere
2×2,687 Posts 
(my emphasis)
Quote:


20200824, 18:17  #7 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2^{5}×3×101 Posts 

Thread Tools  
Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Poker and reasoning.  jwaltos  Probability & Probabilistic Number Theory  20  20191012 03:03 