Do Odd Perfect Numbers Exist? [closed] Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Are there infinitely many Mersenne primes?can't find the proof for the form of odd perfect numbersRelationship between Mersenne Primes and Triangular / Perfect NumbersA question on (odd) perfect numbersWhat is known about multi-perfect numbers?Is it conjectured that there are no odd multi-perfect numbers?Has it been proved that odd perfect numbers cannot be triangular?Odd Perfect number does not existCould a Mersenne prime divide an odd perfect number?In a perfect number $2^p−1 times (2^p − 1)$, the ratio of $p$ to the digits in its perfect number approaches $log(10) / log(4)$?
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Do Odd Perfect Numbers Exist? [closed]
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Are there infinitely many Mersenne primes?can't find the proof for the form of odd perfect numbersRelationship between Mersenne Primes and Triangular / Perfect NumbersA question on (odd) perfect numbersWhat is known about multi-perfect numbers?Is it conjectured that there are no odd multi-perfect numbers?Has it been proved that odd perfect numbers cannot be triangular?Odd Perfect number does not existCould a Mersenne prime divide an odd perfect number?In a perfect number $2^p−1 times (2^p − 1)$, the ratio of $p$ to the digits in its perfect number approaches $log(10) / log(4)$?
$begingroup$
ASSUMPTION: Odd perfect numbers do not exist.
NEED-TO-KNOW:
The perfect numbers are connected to Mersenne Primes in the way that $M_n(M_n+1)/2=P$, where $M_n$ is the $n$th Mersenne Prime and $P$ is the corresponding perfect number. Mersenne Primes are defined as any prime in the form $2^x-1$. However, the $x$ was proven to have to be a prime. Rewriting the Mersenne Prime as $M_p$, where $p$ is the prime that is used to get a Mersenne Prime. This allows a simple conversion formula of $M_p=2^p-1$
PROOF:
First thing to do is to plug the conversion formula into the perfect number formula to get $(2^p-1)(2^p-1+1)/2=P$, which can be simplified to $2^p(2^p-1)/2=P$. Distributing gets $(2^2p-2^p)/2=P$. Notice that you can take away the denominator by subtracting one from both exponents, which gives the final answer $2^2p-1-2^p-1=P$. Since the only way you can get a power of two to be an odd number is by setting the exponent to $0$, but since $2$ is the smallest prime and $2^2-1=2^1$, there is no way that the resulting perfect number could be odd.
EDIT: after hearing what you say and researching the actual equation, I found out that it is really only for even perfects. Proof is wrong.
proof-writing perfect-numbers mersenne-numbers
asked Apr 1 at 17:45
creepercraft97T3creepercraft97T3
104
$endgroup$
closed as unclear what you're asking by Lord Shark the Unknown, Hans Lundmark, Randall, Clayton, Jeremy Rickard Apr 12 at 8:56
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
ASSUMPTION: Odd perfect numbers do not exist.
NEED-TO-KNOW:
The perfect numbers are connected to Mersenne Primes in the way that $M_n(M_n+1)/2=P$, where $M_n$ is the $n$th Mersenne Prime and $P$ is the corresponding perfect number. Mersenne Primes are defined as any prime in the form $2^x-1$. However, the $x$ was proven to have to be a prime. Rewriting the Mersenne Prime as $M_p$, where $p$ is the prime that is used to get a Mersenne Prime. This allows a simple conversion formula of $M_p=2^p-1$
PROOF:
First thing to do is to plug the conversion formula into the perfect number formula to get $(2^p-1)(2^p-1+1)/2=P$, which can be simplified to $2^p(2^p-1)/2=P$. Distributing gets $(2^2p-2^p)/2=P$. Notice that you can take away the denominator by subtracting one from both exponents, which gives the final answer $2^2p-1-2^p-1=P$. Since the only way you can get a power of two to be an odd number is by setting the exponent to $0$, but since $2$ is the smallest prime and $2^2-1=2^1$, there is no way that the resulting perfect number could be odd.
EDIT: after hearing what you say and researching the actual equation, I found out that it is really only for even perfects. Proof is wrong.
proof-writing perfect-numbers mersenne-numbers
asked Apr 1 at 17:45
creepercraft97T3creepercraft97T3
104
$endgroup$
closed as unclear what you're asking by Lord Shark the Unknown, Hans Lundmark, Randall, Clayton, Jeremy Rickard Apr 12 at 8:56
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
Euler proved that held only for even perfect numbers, not all perfect numbers.
$endgroup$
– Randall
Apr 1 at 17:47
$begingroup$
Is the question "is this proof correct?"
$endgroup$
– Randall
Apr 1 at 17:57
add a comment |
$begingroup$
ASSUMPTION: Odd perfect numbers do not exist.
NEED-TO-KNOW:
The perfect numbers are connected to Mersenne Primes in the way that $M_n(M_n+1)/2=P$, where $M_n$ is the $n$th Mersenne Prime and $P$ is the corresponding perfect number. Mersenne Primes are defined as any prime in the form $2^x-1$. However, the $x$ was proven to have to be a prime. Rewriting the Mersenne Prime as $M_p$, where $p$ is the prime that is used to get a Mersenne Prime. This allows a simple conversion formula of $M_p=2^p-1$
PROOF:
First thing to do is to plug the conversion formula into the perfect number formula to get $(2^p-1)(2^p-1+1)/2=P$, which can be simplified to $2^p(2^p-1)/2=P$. Distributing gets $(2^2p-2^p)/2=P$. Notice that you can take away the denominator by subtracting one from both exponents, which gives the final answer $2^2p-1-2^p-1=P$. Since the only way you can get a power of two to be an odd number is by setting the exponent to $0$, but since $2$ is the smallest prime and $2^2-1=2^1$, there is no way that the resulting perfect number could be odd.
EDIT: after hearing what you say and researching the actual equation, I found out that it is really only for even perfects. Proof is wrong.
proof-writing perfect-numbers mersenne-numbers
asked Apr 1 at 17:45
creepercraft97T3creepercraft97T3
104
$endgroup$
ASSUMPTION: Odd perfect numbers do not exist.
NEED-TO-KNOW:
The perfect numbers are connected to Mersenne Primes in the way that $M_n(M_n+1)/2=P$, where $M_n$ is the $n$th Mersenne Prime and $P$ is the corresponding perfect number. Mersenne Primes are defined as any prime in the form $2^x-1$. However, the $x$ was proven to have to be a prime. Rewriting the Mersenne Prime as $M_p$, where $p$ is the prime that is used to get a Mersenne Prime. This allows a simple conversion formula of $M_p=2^p-1$
PROOF:
First thing to do is to plug the conversion formula into the perfect number formula to get $(2^p-1)(2^p-1+1)/2=P$, which can be simplified to $2^p(2^p-1)/2=P$. Distributing gets $(2^2p-2^p)/2=P$. Notice that you can take away the denominator by subtracting one from both exponents, which gives the final answer $2^2p-1-2^p-1=P$. Since the only way you can get a power of two to be an odd number is by setting the exponent to $0$, but since $2$ is the smallest prime and $2^2-1=2^1$, there is no way that the resulting perfect number could be odd.
EDIT: after hearing what you say and researching the actual equation, I found out that it is really only for even perfects. Proof is wrong.
proof-writing perfect-numbers mersenne-numbers
proof-writing perfect-numbers mersenne-numbers
asked Apr 1 at 17:45
creepercraft97T3creepercraft97T3
104
asked Apr 1 at 17:45
creepercraft97T3creepercraft97T3
104
edited Apr 5 at 11:05
creepercraft97T3
asked Apr 1 at 17:45
creepercraft97T3creepercraft97T3
104
asked Apr 1 at 17:45
creepercraft97T3creepercraft97T3
104
asked Apr 1 at 17:45
creepercraft97T3creepercraft97T3
104
104
closed as unclear what you're asking by Lord Shark the Unknown, Hans Lundmark, Randall, Clayton, Jeremy Rickard Apr 12 at 8:56
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by Lord Shark the Unknown, Hans Lundmark, Randall, Clayton, Jeremy Rickard Apr 12 at 8:56
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
Euler proved that held only for even perfect numbers, not all perfect numbers.
$endgroup$
– Randall
Apr 1 at 17:47
$begingroup$
Is the question "is this proof correct?"
$endgroup$
– Randall
Apr 1 at 17:57
add a comment |
1
$begingroup$
Euler proved that held only for even perfect numbers, not all perfect numbers.
$endgroup$
– Randall
Apr 1 at 17:47
$begingroup$
Is the question "is this proof correct?"
$endgroup$
– Randall
Apr 1 at 17:57
1
1
$begingroup$
Euler proved that held only for even perfect numbers, not all perfect numbers.
$endgroup$
– Randall
Apr 1 at 17:47
$begingroup$
Euler proved that held only for even perfect numbers, not all perfect numbers.
$endgroup$
– Randall
Apr 1 at 17:47
$begingroup$
Is the question "is this proof correct?"
$endgroup$
– Randall
Apr 1 at 17:57
$begingroup$
Is the question "is this proof correct?"
$endgroup$
– Randall
Apr 1 at 17:57
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You have correctly quoted the theorem that finds all the even perfect numbers - they come from Mersenne primes. (No one knows whether there are infinitely many but we suspect there are.)
The argument that proves that theorem says nothing about whether there are any odd perfect numbers. No one knows. If there are any they are very big - greater than $10^1500$.
http://mathworld.wolfram.com/OddPerfectNumber.html
answered Apr 1 at 17:48
Ethan BolkerEthan Bolker
46.3k555121
$endgroup$
2
$begingroup$
To add on to this, we've found a lot of very restrictive conditions on the structure of an odd perfect number if it exists. But we haven't ruled out their existence entirely.
$endgroup$
– Don Thousand
Apr 1 at 18:09
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You have correctly quoted the theorem that finds all the even perfect numbers - they come from Mersenne primes. (No one knows whether there are infinitely many but we suspect there are.)
The argument that proves that theorem says nothing about whether there are any odd perfect numbers. No one knows. If there are any they are very big - greater than $10^1500$.
http://mathworld.wolfram.com/OddPerfectNumber.html
answered Apr 1 at 17:48
Ethan BolkerEthan Bolker
46.3k555121
$endgroup$
2
$begingroup$
To add on to this, we've found a lot of very restrictive conditions on the structure of an odd perfect number if it exists. But we haven't ruled out their existence entirely.
$endgroup$
– Don Thousand
Apr 1 at 18:09
add a comment |
$begingroup$
You have correctly quoted the theorem that finds all the even perfect numbers - they come from Mersenne primes. (No one knows whether there are infinitely many but we suspect there are.)
The argument that proves that theorem says nothing about whether there are any odd perfect numbers. No one knows. If there are any they are very big - greater than $10^1500$.
http://mathworld.wolfram.com/OddPerfectNumber.html
answered Apr 1 at 17:48
Ethan BolkerEthan Bolker
46.3k555121
$endgroup$
2
$begingroup$
To add on to this, we've found a lot of very restrictive conditions on the structure of an odd perfect number if it exists. But we haven't ruled out their existence entirely.
$endgroup$
– Don Thousand
Apr 1 at 18:09
add a comment |
$begingroup$
You have correctly quoted the theorem that finds all the even perfect numbers - they come from Mersenne primes. (No one knows whether there are infinitely many but we suspect there are.)
The argument that proves that theorem says nothing about whether there are any odd perfect numbers. No one knows. If there are any they are very big - greater than $10^1500$.
http://mathworld.wolfram.com/OddPerfectNumber.html
answered Apr 1 at 17:48
Ethan BolkerEthan Bolker
46.3k555121
$endgroup$
You have correctly quoted the theorem that finds all the even perfect numbers - they come from Mersenne primes. (No one knows whether there are infinitely many but we suspect there are.)
The argument that proves that theorem says nothing about whether there are any odd perfect numbers. No one knows. If there are any they are very big - greater than $10^1500$.
http://mathworld.wolfram.com/OddPerfectNumber.html
answered Apr 1 at 17:48
Ethan BolkerEthan Bolker
46.3k555121
answered Apr 1 at 17:48
Ethan BolkerEthan Bolker
46.3k555121
answered Apr 1 at 17:48
Ethan BolkerEthan Bolker
46.3k555121
answered Apr 1 at 17:48
Ethan BolkerEthan Bolker
46.3k555121
46.3k555121
2
$begingroup$
To add on to this, we've found a lot of very restrictive conditions on the structure of an odd perfect number if it exists. But we haven't ruled out their existence entirely.
$endgroup$
– Don Thousand
Apr 1 at 18:09
add a comment |
2
$begingroup$
To add on to this, we've found a lot of very restrictive conditions on the structure of an odd perfect number if it exists. But we haven't ruled out their existence entirely.
$endgroup$
– Don Thousand
Apr 1 at 18:09
2
2
$begingroup$
To add on to this, we've found a lot of very restrictive conditions on the structure of an odd perfect number if it exists. But we haven't ruled out their existence entirely.
$endgroup$
– Don Thousand
Apr 1 at 18:09
$begingroup$
To add on to this, we've found a lot of very restrictive conditions on the structure of an odd perfect number if it exists. But we haven't ruled out their existence entirely.
$endgroup$
– Don Thousand
Apr 1 at 18:09
add a comment |
1
$begingroup$
Euler proved that held only for even perfect numbers, not all perfect numbers.
$endgroup$
– Randall
Apr 1 at 17:47
$begingroup$
Is the question "is this proof correct?"
$endgroup$
– Randall
Apr 1 at 17:57