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Error in Twin Prime Conjecture
Can the twin prime conjecture be solved in this way?What is wrong with this proposed proof of the twin prime conjecture?How to pigeonhole the primes between $p_n$ and $p_{n+1}^2$ for twin prime conjecture?Possible method to prove infinite twin prime conjectureTwin prime conjecture proof errorWhy can the sieve of eratosthenes not be used to confirm the twin primes conjecture?Can't understand the logical structure of Euclid's infinitely many primes proof in Rosen's book.A twin prime theorem, and a reformulation of the twin prime conjectureTwin prime conjecture and gaps between primesIterated Twin Prime conjecture
$begingroup$
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
edited 21 mins ago
David G. Stork
11.1k41432
asked 3 hours ago
Jeffrey ScottJeffrey Scott
111
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.
$endgroup$
add a comment |
$begingroup$
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
edited 21 mins ago
David G. Stork
11.1k41432
asked 3 hours ago
Jeffrey ScottJeffrey Scott
111
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
3 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
3 hours ago
add a comment |
$begingroup$
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
edited 21 mins ago
David G. Stork
11.1k41432
asked 3 hours ago
Jeffrey ScottJeffrey Scott
111
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
edited 21 mins ago
David G. Stork
11.1k41432
asked 3 hours ago
Jeffrey ScottJeffrey Scott
111
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.
edited 21 mins ago
David G. Stork
11.1k41432
asked 3 hours ago
Jeffrey ScottJeffrey Scott
111
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.
edited 21 mins ago
David G. Stork
11.1k41432
edited 21 mins ago
David G. Stork
11.1k41432
edited 21 mins ago
David G. Stork
11.1k41432
11.1k41432
asked 3 hours ago
Jeffrey ScottJeffrey Scott
111
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 3 hours ago
Jeffrey ScottJeffrey Scott
111
asked 3 hours ago
Jeffrey ScottJeffrey Scott
111
111
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
3 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
3 hours ago
add a comment |
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
3 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
3 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
3 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
3 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
3 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
3 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered 3 hours ago
Robert ShoreRobert Shore
2,960218
$endgroup$
2
$begingroup$
Ah you're right. 2 * 3 * 5 * 7 = 210. But 209 is not prime.
$endgroup$
– Jeffrey Scott
3 hours ago
$begingroup$
Glad I could help. Acceptances of answers that you find useful are always welcome.
$endgroup$
– Robert Shore
2 hours ago
add a comment |
Your Answer
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1 Answer
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$begingroup$
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.
answered 3 hours ago
Robert ShoreRobert Shore
2,960218
$endgroup$
2
$begingroup$
Ah you're right. 2 * 3 * 5 * 7 = 210. But 209 is not prime.
$endgroup$
– Jeffrey Scott
3 hours ago
$begingroup$
Glad I could help. Acceptances of answers that you find useful are always welcome.
$endgroup$
– Robert Shore
2 hours ago
add a comment |
$begingroup$
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.
answered 3 hours ago
Robert ShoreRobert Shore
2,960218
$endgroup$
2
$begingroup$
Ah you're right. 2 * 3 * 5 * 7 = 210. But 209 is not prime.
$endgroup$
– Jeffrey Scott
3 hours ago
$begingroup$
Glad I could help. Acceptances of answers that you find useful are always welcome.
$endgroup$
– Robert Shore
2 hours ago
add a comment |
$begingroup$
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.
answered 3 hours ago
Robert ShoreRobert Shore
2,960218
$endgroup$
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.
answered 3 hours ago
Robert ShoreRobert Shore
2,960218
answered 3 hours ago
Robert ShoreRobert Shore
2,960218
answered 3 hours ago
Robert ShoreRobert Shore
2,960218
answered 3 hours ago
Robert ShoreRobert Shore
2,960218
2,960218
2
$begingroup$
Ah you're right. 2 * 3 * 5 * 7 = 210. But 209 is not prime.
$endgroup$
– Jeffrey Scott
3 hours ago
$begingroup$
Glad I could help. Acceptances of answers that you find useful are always welcome.
$endgroup$
– Robert Shore
2 hours ago
add a comment |
2
$begingroup$
Ah you're right. 2 * 3 * 5 * 7 = 210. But 209 is not prime.
$endgroup$
– Jeffrey Scott
3 hours ago
$begingroup$
Glad I could help. Acceptances of answers that you find useful are always welcome.
$endgroup$
– Robert Shore
2 hours ago
2
2
$begingroup$
Ah you're right. 2 * 3 * 5 * 7 = 210. But 209 is not prime.
$endgroup$
– Jeffrey Scott
3 hours ago
$begingroup$
Ah you're right. 2 * 3 * 5 * 7 = 210. But 209 is not prime.
$endgroup$
– Jeffrey Scott
3 hours ago
$begingroup$
Glad I could help. Acceptances of answers that you find useful are always welcome.
$endgroup$
– Robert Shore
2 hours ago
$begingroup$
Glad I could help. Acceptances of answers that you find useful are always welcome.
$endgroup$
– Robert Shore
2 hours ago
add a comment |
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.
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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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
3 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
3 hours ago