20190415, 13:55  #23  
Undefined
"The unspeakable one"
Jun 2006
My evil lair
2·3·5·193 Posts 
Quote:
And if 2 isn't a twin prime then 3×5+1 = 2×2×2×2 Either way there are always possible divisors that aren't in the original set of primes you multiplied together. Last fiddled with by retina on 20190415 at 13:55 

20190415, 14:05  #24  
"Curtis"
Feb 2005
Riverside, CA
10444_{8} Posts 
Quote:
Your Q is divisible by 2 (do you see why?). All other claims about properties of Q's factors must be proven. Prove your claim. 

20190415, 14:46  #25 
Mar 2019
5×11 Posts 
I will try to fix the proof once again
Assume that there exist a finite amount of twin prime numbers. Then we can construct a list, which in this case will be S, S= A1,A2,A3,A4...An Let P be the product of all twin prime numbers in S, P= A1*A2*A3*A4..An Let Q=P+1 If Q is a twin prime number then S is not complete If Q is composite then some prime factor p divides Q, if this factor p were in our list S then it would divide P, but p divides P+1=Q. If p divides P and Q then p would have to divide the difference of the two numbers, which is (P+1)P or just 1. Since no twin prime number divides 1, p can not be on the list. This means that at least one twin prime number exists beyond those in the list. Last fiddled with by MathDoggy on 20190415 at 15:30 
20190415, 14:55  #26  
Undefined
"The unspeakable one"
Jun 2006
My evil lair
2·3·5·193 Posts 
Quote:
Also, p doesn't have to be a twin prime, it can be some other prime, like 2 for example. 

20190415, 15:03  #27  
"William"
May 2003
New Haven
2^{2}·3^{2}·5·13 Posts 
Quote:
Ooops. This means at least one prime (p) exists beyond those in the list. But why must it be a twin prime? 

20190415, 15:03  #28 
Mar 2019
5·11 Posts 
What do I have to fix, retina

20190415, 15:16  #29 
Mar 2019
5×11 Posts 

20190415, 15:43  #30 
Mar 2019
5·11 Posts 
Ooops. This means at least one prime (p) exists beyond those in the list. But why must it be a twin prime?[/QUOTE]
There is no such implication 
20190415, 16:02  #31 
Mar 2019
37_{16} Posts 
Another attempt with an alternate method
Just ignore this
Last fiddled with by MathDoggy on 20190415 at 16:13 
20190415, 16:11  #32  
Aug 2006
2^{3}·3·13·19 Posts 
Quote:
Last fiddled with by CRGreathouse on 20190415 at 16:18 

20190415, 16:22  #33 
Mar 2019
5·11 Posts 
Should I quit to try to fix the proof?
It is because nothing is working Last fiddled with by MathDoggy on 20190415 at 16:23 
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