What is the state of Carmichael's totient function conjecture?
$begingroup$
I have been searching for information about that conjecture and it seems for me that noone has made any significant improvement on it in the last 30 years.
Is that true? Does it remain unproven to be true? Has there been any important discovery about the problem in the last years?
Thank you.
number-theory elementary-number-theory totient-function conjectures carmichael-function
asked Oct 16 '16 at 21:01
user3141592user3141592
894623
$endgroup$
add a comment |
$begingroup$
I have been searching for information about that conjecture and it seems for me that noone has made any significant improvement on it in the last 30 years.
Is that true? Does it remain unproven to be true? Has there been any important discovery about the problem in the last years?
Thank you.
number-theory elementary-number-theory totient-function conjectures carmichael-function
asked Oct 16 '16 at 21:01
user3141592user3141592
894623
$endgroup$
1
$begingroup$
Are you asking about Carmichael's totient conjecture or Lehmer's? The title refers to Carmichael, but the preprint you link to is exclusively about Lehmer. These are two entirely different conjectures (Carmichael is about the multiplicity of values taken by $phi$, and Lehmer is about the solubility of $phi(n) mid n-1$).
$endgroup$
– Erick Wong
Oct 16 '16 at 22:17
$begingroup$
If you really want to know about Carmichael's conjecture, then it is obvious from the Wikipedia article that Kevin Ford has significantly advanced our understanding of it less than 20 years ago. So I repeat, do you really mean Carmichael's conjecture?
$endgroup$
– Erick Wong
Feb 24 '17 at 7:16
add a comment |
$begingroup$
I have been searching for information about that conjecture and it seems for me that noone has made any significant improvement on it in the last 30 years.
Is that true? Does it remain unproven to be true? Has there been any important discovery about the problem in the last years?
Thank you.
number-theory elementary-number-theory totient-function conjectures carmichael-function
asked Oct 16 '16 at 21:01
user3141592user3141592
894623
$endgroup$
I have been searching for information about that conjecture and it seems for me that noone has made any significant improvement on it in the last 30 years.
Is that true? Does it remain unproven to be true? Has there been any important discovery about the problem in the last years?
Thank you.
number-theory elementary-number-theory totient-function conjectures carmichael-function
number-theory elementary-number-theory totient-function conjectures carmichael-function
asked Oct 16 '16 at 21:01
user3141592user3141592
894623
asked Oct 16 '16 at 21:01
user3141592user3141592
894623
edited Feb 24 '17 at 6:04
user3141592
asked Oct 16 '16 at 21:01
user3141592user3141592
894623
asked Oct 16 '16 at 21:01
user3141592user3141592
894623
asked Oct 16 '16 at 21:01
user3141592user3141592
894623
894623
1
$begingroup$
Are you asking about Carmichael's totient conjecture or Lehmer's? The title refers to Carmichael, but the preprint you link to is exclusively about Lehmer. These are two entirely different conjectures (Carmichael is about the multiplicity of values taken by $phi$, and Lehmer is about the solubility of $phi(n) mid n-1$).
$endgroup$
– Erick Wong
Oct 16 '16 at 22:17
$begingroup$
If you really want to know about Carmichael's conjecture, then it is obvious from the Wikipedia article that Kevin Ford has significantly advanced our understanding of it less than 20 years ago. So I repeat, do you really mean Carmichael's conjecture?
$endgroup$
– Erick Wong
Feb 24 '17 at 7:16
add a comment |
1
$begingroup$
Are you asking about Carmichael's totient conjecture or Lehmer's? The title refers to Carmichael, but the preprint you link to is exclusively about Lehmer. These are two entirely different conjectures (Carmichael is about the multiplicity of values taken by $phi$, and Lehmer is about the solubility of $phi(n) mid n-1$).
$endgroup$
– Erick Wong
Oct 16 '16 at 22:17
$begingroup$
If you really want to know about Carmichael's conjecture, then it is obvious from the Wikipedia article that Kevin Ford has significantly advanced our understanding of it less than 20 years ago. So I repeat, do you really mean Carmichael's conjecture?
$endgroup$
– Erick Wong
Feb 24 '17 at 7:16
1
1
$begingroup$
Are you asking about Carmichael's totient conjecture or Lehmer's? The title refers to Carmichael, but the preprint you link to is exclusively about Lehmer. These are two entirely different conjectures (Carmichael is about the multiplicity of values taken by $phi$, and Lehmer is about the solubility of $phi(n) mid n-1$).
$endgroup$
– Erick Wong
Oct 16 '16 at 22:17
$begingroup$
Are you asking about Carmichael's totient conjecture or Lehmer's? The title refers to Carmichael, but the preprint you link to is exclusively about Lehmer. These are two entirely different conjectures (Carmichael is about the multiplicity of values taken by $phi$, and Lehmer is about the solubility of $phi(n) mid n-1$).
$endgroup$
– Erick Wong
Oct 16 '16 at 22:17
$begingroup$
If you really want to know about Carmichael's conjecture, then it is obvious from the Wikipedia article that Kevin Ford has significantly advanced our understanding of it less than 20 years ago. So I repeat, do you really mean Carmichael's conjecture?
$endgroup$
– Erick Wong
Feb 24 '17 at 7:16
$begingroup$
If you really want to know about Carmichael's conjecture, then it is obvious from the Wikipedia article that Kevin Ford has significantly advanced our understanding of it less than 20 years ago. So I repeat, do you really mean Carmichael's conjecture?
$endgroup$
– Erick Wong
Feb 24 '17 at 7:16
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let's define Carmichael's Totient Conjecture:
For each $n$, there exists an integer $mneq n$ such that $φ(m) = φ(n) = k$. Where $φ$ defined to be Euler's Totient.
The conjecture is an open problem in general, but is proven for all $k$ such that $k+1$ is prime.
Proof: Suppose $n$ is prime and $n-1 = k$. Then $φ(n) = k$. Now $φ(2) = 1$, and the totient of any integer $t$ is the product of totients of primes powers dividing $t$. Now let $m = 2n$. Since the only prime powers dividing $m$ are $2$ and $n$, $φ(m) = (2-1)*(n-1)$ $=$ $φ(m) = k$, therefore $φ(m) = φ(n) = k$.
edited Jan 1 at 14:40
Giovanni Di Matteo
7816
answered Feb 24 '17 at 6:14
J. LinneJ. Linne
888415
$endgroup$
1
$begingroup$
More generally this works for any odd $n$ - $2n$ then has the same value of the totient function.
$endgroup$
– Wojowu
Feb 24 '17 at 6:34
$begingroup$
@GottfriedHelms What about all the numbers with higher power of 2 in them?
$endgroup$
– Wojowu
Feb 24 '17 at 7:17
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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$begingroup$
Let's define Carmichael's Totient Conjecture:
For each $n$, there exists an integer $mneq n$ such that $φ(m) = φ(n) = k$. Where $φ$ defined to be Euler's Totient.
The conjecture is an open problem in general, but is proven for all $k$ such that $k+1$ is prime.
Proof: Suppose $n$ is prime and $n-1 = k$. Then $φ(n) = k$. Now $φ(2) = 1$, and the totient of any integer $t$ is the product of totients of primes powers dividing $t$. Now let $m = 2n$. Since the only prime powers dividing $m$ are $2$ and $n$, $φ(m) = (2-1)*(n-1)$ $=$ $φ(m) = k$, therefore $φ(m) = φ(n) = k$.
edited Jan 1 at 14:40
Giovanni Di Matteo
7816
answered Feb 24 '17 at 6:14
J. LinneJ. Linne
888415
$endgroup$
1
$begingroup$
More generally this works for any odd $n$ - $2n$ then has the same value of the totient function.
$endgroup$
– Wojowu
Feb 24 '17 at 6:34
$begingroup$
@GottfriedHelms What about all the numbers with higher power of 2 in them?
$endgroup$
– Wojowu
Feb 24 '17 at 7:17
add a comment |
$begingroup$
Let's define Carmichael's Totient Conjecture:
For each $n$, there exists an integer $mneq n$ such that $φ(m) = φ(n) = k$. Where $φ$ defined to be Euler's Totient.
The conjecture is an open problem in general, but is proven for all $k$ such that $k+1$ is prime.
Proof: Suppose $n$ is prime and $n-1 = k$. Then $φ(n) = k$. Now $φ(2) = 1$, and the totient of any integer $t$ is the product of totients of primes powers dividing $t$. Now let $m = 2n$. Since the only prime powers dividing $m$ are $2$ and $n$, $φ(m) = (2-1)*(n-1)$ $=$ $φ(m) = k$, therefore $φ(m) = φ(n) = k$.
edited Jan 1 at 14:40
Giovanni Di Matteo
7816
answered Feb 24 '17 at 6:14
J. LinneJ. Linne
888415
$endgroup$
1
$begingroup$
More generally this works for any odd $n$ - $2n$ then has the same value of the totient function.
$endgroup$
– Wojowu
Feb 24 '17 at 6:34
$begingroup$
@GottfriedHelms What about all the numbers with higher power of 2 in them?
$endgroup$
– Wojowu
Feb 24 '17 at 7:17
add a comment |
$begingroup$
Let's define Carmichael's Totient Conjecture:
For each $n$, there exists an integer $mneq n$ such that $φ(m) = φ(n) = k$. Where $φ$ defined to be Euler's Totient.
The conjecture is an open problem in general, but is proven for all $k$ such that $k+1$ is prime.
Proof: Suppose $n$ is prime and $n-1 = k$. Then $φ(n) = k$. Now $φ(2) = 1$, and the totient of any integer $t$ is the product of totients of primes powers dividing $t$. Now let $m = 2n$. Since the only prime powers dividing $m$ are $2$ and $n$, $φ(m) = (2-1)*(n-1)$ $=$ $φ(m) = k$, therefore $φ(m) = φ(n) = k$.
edited Jan 1 at 14:40
Giovanni Di Matteo
7816
answered Feb 24 '17 at 6:14
J. LinneJ. Linne
888415
$endgroup$
Let's define Carmichael's Totient Conjecture:
For each $n$, there exists an integer $mneq n$ such that $φ(m) = φ(n) = k$. Where $φ$ defined to be Euler's Totient.
The conjecture is an open problem in general, but is proven for all $k$ such that $k+1$ is prime.
Proof: Suppose $n$ is prime and $n-1 = k$. Then $φ(n) = k$. Now $φ(2) = 1$, and the totient of any integer $t$ is the product of totients of primes powers dividing $t$. Now let $m = 2n$. Since the only prime powers dividing $m$ are $2$ and $n$, $φ(m) = (2-1)*(n-1)$ $=$ $φ(m) = k$, therefore $φ(m) = φ(n) = k$.
edited Jan 1 at 14:40
Giovanni Di Matteo
7816
answered Feb 24 '17 at 6:14
J. LinneJ. Linne
888415
edited Jan 1 at 14:40
Giovanni Di Matteo
7816
edited Jan 1 at 14:40
Giovanni Di Matteo
7816
edited Jan 1 at 14:40
Giovanni Di Matteo
7816
7816
answered Feb 24 '17 at 6:14
J. LinneJ. Linne
888415
answered Feb 24 '17 at 6:14
J. LinneJ. Linne
888415
answered Feb 24 '17 at 6:14
J. LinneJ. Linne
888415
888415
1
$begingroup$
More generally this works for any odd $n$ - $2n$ then has the same value of the totient function.
$endgroup$
– Wojowu
Feb 24 '17 at 6:34
$begingroup$
@GottfriedHelms What about all the numbers with higher power of 2 in them?
$endgroup$
– Wojowu
Feb 24 '17 at 7:17
add a comment |
1
$begingroup$
More generally this works for any odd $n$ - $2n$ then has the same value of the totient function.
$endgroup$
– Wojowu
Feb 24 '17 at 6:34
$begingroup$
@GottfriedHelms What about all the numbers with higher power of 2 in them?
$endgroup$
– Wojowu
Feb 24 '17 at 7:17
1
1
$begingroup$
More generally this works for any odd $n$ - $2n$ then has the same value of the totient function.
$endgroup$
– Wojowu
Feb 24 '17 at 6:34
$begingroup$
More generally this works for any odd $n$ - $2n$ then has the same value of the totient function.
$endgroup$
– Wojowu
Feb 24 '17 at 6:34
$begingroup$
@GottfriedHelms What about all the numbers with higher power of 2 in them?
$endgroup$
– Wojowu
Feb 24 '17 at 7:17
$begingroup$
@GottfriedHelms What about all the numbers with higher power of 2 in them?
$endgroup$
– Wojowu
Feb 24 '17 at 7:17
add a comment |
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1
$begingroup$
Are you asking about Carmichael's totient conjecture or Lehmer's? The title refers to Carmichael, but the preprint you link to is exclusively about Lehmer. These are two entirely different conjectures (Carmichael is about the multiplicity of values taken by $phi$, and Lehmer is about the solubility of $phi(n) mid n-1$).
$endgroup$
– Erick Wong
Oct 16 '16 at 22:17
$begingroup$
If you really want to know about Carmichael's conjecture, then it is obvious from the Wikipedia article that Kevin Ford has significantly advanced our understanding of it less than 20 years ago. So I repeat, do you really mean Carmichael's conjecture?
$endgroup$
– Erick Wong
Feb 24 '17 at 7:16