mersenneforum.org  

Go Back   mersenneforum.org > Extra Stuff > Miscellaneous Math

Reply
 
Thread Tools
Old 2020-08-23, 10:50   #1
Charles Kusniec
 
Charles Kusniec's Avatar
 
Aug 2020
Brasil

2·3 Posts
Default 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=r-d

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,

P1-P2=2d

d=(P1-P2)/2.

If r=Odd, then d=Even and vice-versa.

OK.

Now, the Semiprime formed with the two Primes above is

P1P2=(r+d)(r-d)

So,

P1P2=r^2-d^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.
Charles Kusniec is offline   Reply With Quote
Old 2020-08-23, 12:42   #2
Viliam Furik
 
Viliam Furik's Avatar
 
"Viliam Furík"
Jul 2018
Martin, Slovakia

2×11×31 Posts
Default

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.
Viliam Furik is offline   Reply With Quote
Old 2020-08-23, 12:50   #3
kruoli
 
kruoli's Avatar
 
"Oliver"
Sep 2017
Porta Westfalica, DE

29A16 Posts
Default

Quote:
Originally Posted by Viliam Furik View Post
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
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.
kruoli is offline   Reply With Quote
Old 2020-08-24, 09:24   #4
LaurV
Romulan Interpreter
 
LaurV's Avatar
 
"name field"
Jun 2011
Thailand

9,787 Posts
Default

Quote:
Originally Posted by Charles Kusniec View Post
Where is the weakness of this reasoning?
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 2020-08-24 at 09:34
LaurV is offline   Reply With Quote
Old 2020-08-24, 09:40   #5
LaurV
Romulan Interpreter
 
LaurV's Avatar
 
"name field"
Jun 2011
Thailand

9,787 Posts
Default

Quote:
Originally Posted by Viliam Furik View Post
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.
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.
LaurV is offline   Reply With Quote
Old 2020-08-24, 12:11   #6
Dr Sardonicus
 
Dr Sardonicus's Avatar
 
Feb 2017
Nowhere

499710 Posts
Default

(my emphasis)
Quote:
Originally Posted by Charles Kusniec View Post
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.
That is what may be called "too big an if."
Dr Sardonicus is offline   Reply With Quote
Old 2020-08-24, 18:17   #7
Batalov
 
Batalov's Avatar
 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2

957110 Posts
Default

Quote:
Originally Posted by Charles Kusniec View Post
if any Odd number can be expressed as a difference of two Squares
Of course it can! 2n + 1 = (n+1)2 - n2
Batalov is offline   Reply With Quote
Reply

Thread Tools


Similar Threads
Thread Thread Starter Forum Replies Last Post
Poker and reasoning. jwaltos Probability & Probabilistic Number Theory 20 2019-10-12 03:03

All times are UTC. The time now is 15:42.


Sun Oct 24 15:42:17 UTC 2021 up 93 days, 10:11, 0 users, load averages: 1.87, 2.57, 2.38

Powered by vBulletin® Version 3.8.11
Copyright ©2000 - 2021, Jelsoft Enterprises Ltd.

This forum has received and complied with 0 (zero) government requests for information.

Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation.
A copy of the license is included in the FAQ.