Error in Twin Prime Conjecture












3












$begingroup$


Euclid's theorem states:




Consider any finite list of prime numbers $p_1, p_2, ..., p_n$. It will be shown that at least one additional prime number not in this list exists. Let $P$ be the product of all the prime numbers in the list: $P = p_1p_2...p_n$. Let $q = P + 1$. Then $q$ is either prime or not.



If $q$ is prime, then there is at least one more prime that is not in the list. If $q$ is not prime, then some prime factor $p$ divides $q$. If this factor $p$ were in our list, then it would divide $P$ (since $P$ is the product of every number in the list); 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 prime number divides $1$, $p$ cannot be on the list. This means that at least one more prime number exists beyond those in the list. This proves that for every finite list of prime numbers there is a prime number not in the list, and therefore there must be infinitely many prime numbers.





My question:



Does this theorem also hold if you let $q = P - 1$?



Wouldn't $P-1$ also be necessarily a new prime number? And if so, it and $P+1$ would be a set of twin primes.




So the proof would be:



Assume there are a finite number of twin primes such that $p_{n+1} - p_n = 2$.



Then, from the final set of twin primes, choose the larger of these two primes $p_{n+1}$. Calculate $S=p_1p_2...p_{n+1}$. So you now have a product of all primes up to $p_{n+1}$. Call this $S$. $S + 1$ is a prime number and so is $S - 1$. This is a new set of twin primes not in our original list, thus there cannot be a finite list of twin primes.



Of course, if $S - 1$ is not prime, then this falls apart.






proof-verification prime-numbers prime-twins







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Jeffrey Scott is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • 6




    $begingroup$
    There’s absolutely no reason why S+1 or S-1 should be a prime number though, it merely has an unlisted prime factor.
    $endgroup$
    – Noe Blassel
    4 hours ago










  • $begingroup$
    By the way, take a look at the edits. It's a courtesy to other contributors to use MathJax to format your posts. If you're not familiar with it, it's not hard to learn -- I've been on this site for less than two months and it has become second nature.
    $endgroup$
    – Robert Shore
    4 hours ago
















3












$begingroup$


Euclid's theorem states:




Consider any finite list of prime numbers $p_1, p_2, ..., p_n$. It will be shown that at least one additional prime number not in this list exists. Let $P$ be the product of all the prime numbers in the list: $P = p_1p_2...p_n$. Let $q = P + 1$. Then $q$ is either prime or not.



If $q$ is prime, then there is at least one more prime that is not in the list. If $q$ is not prime, then some prime factor $p$ divides $q$. If this factor $p$ were in our list, then it would divide $P$ (since $P$ is the product of every number in the list); 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 prime number divides $1$, $p$ cannot be on the list. This means that at least one more prime number exists beyond those in the list. This proves that for every finite list of prime numbers there is a prime number not in the list, and therefore there must be infinitely many prime numbers.





My question:



Does this theorem also hold if you let $q = P - 1$?



Wouldn't $P-1$ also be necessarily a new prime number? And if so, it and $P+1$ would be a set of twin primes.




So the proof would be:



Assume there are a finite number of twin primes such that $p_{n+1} - p_n = 2$.



Then, from the final set of twin primes, choose the larger of these two primes $p_{n+1}$. Calculate $S=p_1p_2...p_{n+1}$. So you now have a product of all primes up to $p_{n+1}$. Call this $S$. $S + 1$ is a prime number and so is $S - 1$. This is a new set of twin primes not in our original list, thus there cannot be a finite list of twin primes.



Of course, if $S - 1$ is not prime, then this falls apart.






proof-verification prime-numbers prime-twins







share|cite|improve this question









New contributor




Jeffrey Scott is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 6




    $begingroup$
    There’s absolutely no reason why S+1 or S-1 should be a prime number though, it merely has an unlisted prime factor.
    $endgroup$
    – Noe Blassel
    4 hours ago










  • $begingroup$
    By the way, take a look at the edits. It's a courtesy to other contributors to use MathJax to format your posts. If you're not familiar with it, it's not hard to learn -- I've been on this site for less than two months and it has become second nature.
    $endgroup$
    – Robert Shore
    4 hours ago














3












3








3


1



$begingroup$


Euclid's theorem states:




Consider any finite list of prime numbers $p_1, p_2, ..., p_n$. It will be shown that at least one additional prime number not in this list exists. Let $P$ be the product of all the prime numbers in the list: $P = p_1p_2...p_n$. Let $q = P + 1$. Then $q$ is either prime or not.



If $q$ is prime, then there is at least one more prime that is not in the list. If $q$ is not prime, then some prime factor $p$ divides $q$. If this factor $p$ were in our list, then it would divide $P$ (since $P$ is the product of every number in the list); 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 prime number divides $1$, $p$ cannot be on the list. This means that at least one more prime number exists beyond those in the list. This proves that for every finite list of prime numbers there is a prime number not in the list, and therefore there must be infinitely many prime numbers.





My question:



Does this theorem also hold if you let $q = P - 1$?



Wouldn't $P-1$ also be necessarily a new prime number? And if so, it and $P+1$ would be a set of twin primes.




So the proof would be:



Assume there are a finite number of twin primes such that $p_{n+1} - p_n = 2$.



Then, from the final set of twin primes, choose the larger of these two primes $p_{n+1}$. Calculate $S=p_1p_2...p_{n+1}$. So you now have a product of all primes up to $p_{n+1}$. Call this $S$. $S + 1$ is a prime number and so is $S - 1$. This is a new set of twin primes not in our original list, thus there cannot be a finite list of twin primes.



Of course, if $S - 1$ is not prime, then this falls apart.






proof-verification prime-numbers prime-twins







share|cite|improve this question









New contributor




Jeffrey Scott is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




Euclid's theorem states:




Consider any finite list of prime numbers $p_1, p_2, ..., p_n$. It will be shown that at least one additional prime number not in this list exists. Let $P$ be the product of all the prime numbers in the list: $P = p_1p_2...p_n$. Let $q = P + 1$. Then $q$ is either prime or not.



If $q$ is prime, then there is at least one more prime that is not in the list. If $q$ is not prime, then some prime factor $p$ divides $q$. If this factor $p$ were in our list, then it would divide $P$ (since $P$ is the product of every number in the list); 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 prime number divides $1$, $p$ cannot be on the list. This means that at least one more prime number exists beyond those in the list. This proves that for every finite list of prime numbers there is a prime number not in the list, and therefore there must be infinitely many prime numbers.





My question:



Does this theorem also hold if you let $q = P - 1$?



Wouldn't $P-1$ also be necessarily a new prime number? And if so, it and $P+1$ would be a set of twin primes.




So the proof would be:



Assume there are a finite number of twin primes such that $p_{n+1} - p_n = 2$.



Then, from the final set of twin primes, choose the larger of these two primes $p_{n+1}$. Calculate $S=p_1p_2...p_{n+1}$. So you now have a product of all primes up to $p_{n+1}$. Call this $S$. $S + 1$ is a prime number and so is $S - 1$. This is a new set of twin primes not in our original list, thus there cannot be a finite list of twin primes.



Of course, if $S - 1$ is not prime, then this falls apart.






proof-verification prime-numbers prime-twins




proof-verification prime-numbers prime-twins






share|cite|improve this question









New contributor




Jeffrey Scott is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




Jeffrey Scott is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 1 hour ago









David G. Stork

11.1k41432




11.1k41432






New contributor




Jeffrey Scott is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 4 hours ago









Jeffrey ScottJeffrey Scott

161




161




New contributor




Jeffrey Scott is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Jeffrey Scott is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Jeffrey Scott is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








  • 6




    $begingroup$
    There’s absolutely no reason why S+1 or S-1 should be a prime number though, it merely has an unlisted prime factor.
    $endgroup$
    – Noe Blassel
    4 hours ago










  • $begingroup$
    By the way, take a look at the edits. It's a courtesy to other contributors to use MathJax to format your posts. If you're not familiar with it, it's not hard to learn -- I've been on this site for less than two months and it has become second nature.
    $endgroup$
    – Robert Shore
    4 hours ago














  • 6




    $begingroup$
    There’s absolutely no reason why S+1 or S-1 should be a prime number though, it merely has an unlisted prime factor.
    $endgroup$
    – Noe Blassel
    4 hours ago










  • $begingroup$
    By the way, take a look at the edits. It's a courtesy to other contributors to use MathJax to format your posts. If you're not familiar with it, it's not hard to learn -- I've been on this site for less than two months and it has become second nature.
    $endgroup$
    – Robert Shore
    4 hours ago








6




6




$begingroup$
There’s absolutely no reason why S+1 or S-1 should be a prime number though, it merely has an unlisted prime factor.
$endgroup$
– Noe Blassel
4 hours ago




$begingroup$
There’s absolutely no reason why S+1 or S-1 should be a prime number though, it merely has an unlisted prime factor.
$endgroup$
– Noe Blassel
4 hours ago












$begingroup$
By the way, take a look at the edits. It's a courtesy to other contributors to use MathJax to format your posts. If you're not familiar with it, it's not hard to learn -- I've been on this site for less than two months and it has become second nature.
$endgroup$
– Robert Shore
4 hours ago




$begingroup$
By the way, take a look at the edits. It's a courtesy to other contributors to use MathJax to format your posts. If you're not familiar with it, it's not hard to learn -- I've been on this site for less than two months and it has become second nature.
$endgroup$
– Robert Shore
4 hours ago










1 Answer
1






active

oldest

votes


















7












$begingroup$

Remember, your original argument doesn't show that $P+1$ is itself prime. It shows that $P+1$ has a prime factor that you haven't already accounted for. So while you could make the same argument for $P-1$, you'd also reach the same conclusion, not that $P-1$ is itself necessarily prime, but only that it has some prime factor not in your original list. So that's of no help in proving the Twin Prime Conjecture.



Similarly, you don't know that $S+1$ or $S-1$ is prime. You just know that they have prime factors that aren't on your original list of twin primes, but that doesn't help you.






share|cite|improve this answer









$endgroup$









  • 3




    $begingroup$
    Ah you're right. 2 * 3 * 5 * 7 = 210. But 209 is not prime.
    $endgroup$
    – Jeffrey Scott
    4 hours ago










  • $begingroup$
    Glad I could help. Acceptances of answers that you find useful are always welcome.
    $endgroup$
    – Robert Shore
    3 hours ago











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1 Answer
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7












$begingroup$

Remember, your original argument doesn't show that $P+1$ is itself prime. It shows that $P+1$ has a prime factor that you haven't already accounted for. So while you could make the same argument for $P-1$, you'd also reach the same conclusion, not that $P-1$ is itself necessarily prime, but only that it has some prime factor not in your original list. So that's of no help in proving the Twin Prime Conjecture.



Similarly, you don't know that $S+1$ or $S-1$ is prime. You just know that they have prime factors that aren't on your original list of twin primes, but that doesn't help you.






share|cite|improve this answer









$endgroup$









  • 3




    $begingroup$
    Ah you're right. 2 * 3 * 5 * 7 = 210. But 209 is not prime.
    $endgroup$
    – Jeffrey Scott
    4 hours ago










  • $begingroup$
    Glad I could help. Acceptances of answers that you find useful are always welcome.
    $endgroup$
    – Robert Shore
    3 hours ago
















7












$begingroup$

Remember, your original argument doesn't show that $P+1$ is itself prime. It shows that $P+1$ has a prime factor that you haven't already accounted for. So while you could make the same argument for $P-1$, you'd also reach the same conclusion, not that $P-1$ is itself necessarily prime, but only that it has some prime factor not in your original list. So that's of no help in proving the Twin Prime Conjecture.



Similarly, you don't know that $S+1$ or $S-1$ is prime. You just know that they have prime factors that aren't on your original list of twin primes, but that doesn't help you.






share|cite|improve this answer









$endgroup$









  • 3




    $begingroup$
    Ah you're right. 2 * 3 * 5 * 7 = 210. But 209 is not prime.
    $endgroup$
    – Jeffrey Scott
    4 hours ago










  • $begingroup$
    Glad I could help. Acceptances of answers that you find useful are always welcome.
    $endgroup$
    – Robert Shore
    3 hours ago














7












7








7





$begingroup$

Remember, your original argument doesn't show that $P+1$ is itself prime. It shows that $P+1$ has a prime factor that you haven't already accounted for. So while you could make the same argument for $P-1$, you'd also reach the same conclusion, not that $P-1$ is itself necessarily prime, but only that it has some prime factor not in your original list. So that's of no help in proving the Twin Prime Conjecture.



Similarly, you don't know that $S+1$ or $S-1$ is prime. You just know that they have prime factors that aren't on your original list of twin primes, but that doesn't help you.






share|cite|improve this answer









$endgroup$



Remember, your original argument doesn't show that $P+1$ is itself prime. It shows that $P+1$ has a prime factor that you haven't already accounted for. So while you could make the same argument for $P-1$, you'd also reach the same conclusion, not that $P-1$ is itself necessarily prime, but only that it has some prime factor not in your original list. So that's of no help in proving the Twin Prime Conjecture.



Similarly, you don't know that $S+1$ or $S-1$ is prime. You just know that they have prime factors that aren't on your original list of twin primes, but that doesn't help you.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 4 hours ago









Robert ShoreRobert Shore

2,980219




2,980219








  • 3




    $begingroup$
    Ah you're right. 2 * 3 * 5 * 7 = 210. But 209 is not prime.
    $endgroup$
    – Jeffrey Scott
    4 hours ago










  • $begingroup$
    Glad I could help. Acceptances of answers that you find useful are always welcome.
    $endgroup$
    – Robert Shore
    3 hours ago














  • 3




    $begingroup$
    Ah you're right. 2 * 3 * 5 * 7 = 210. But 209 is not prime.
    $endgroup$
    – Jeffrey Scott
    4 hours ago










  • $begingroup$
    Glad I could help. Acceptances of answers that you find useful are always welcome.
    $endgroup$
    – Robert Shore
    3 hours ago








3




3




$begingroup$
Ah you're right. 2 * 3 * 5 * 7 = 210. But 209 is not prime.
$endgroup$
– Jeffrey Scott
4 hours ago




$begingroup$
Ah you're right. 2 * 3 * 5 * 7 = 210. But 209 is not prime.
$endgroup$
– Jeffrey Scott
4 hours ago












$begingroup$
Glad I could help. Acceptances of answers that you find useful are always welcome.
$endgroup$
– Robert Shore
3 hours ago




$begingroup$
Glad I could help. Acceptances of answers that you find useful are always welcome.
$endgroup$
– Robert Shore
3 hours ago










Jeffrey Scott is a new contributor. Be nice, and check out our Code of Conduct.










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Jeffrey Scott is a new contributor. Be nice, and check out our Code of Conduct.












Jeffrey Scott is a new contributor. Be nice, and check out our Code of Conduct.
















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