How do you find all $n$ such that $phi(n)|n$
Where $phi(n)$ is the Euler phi function, how do you find all $n$ such that $phi(n)|n$?
elementary-number-theory
asked Apr 23 '12 at 12:22
Freeman
88082163
|
show 1 more comment
Where $phi(n)$ is the Euler phi function, how do you find all $n$ such that $phi(n)|n$?
elementary-number-theory
asked Apr 23 '12 at 12:22
Freeman
88082163
4
mathforum.org/kb/… and oeis.org/A007694
– Beni Bogosel
Apr 23 '12 at 12:24
2
Use the formula for the totient function in terms of the prime factors of n.
– Dayo Adeyemi
Apr 23 '12 at 12:32
@dayo Adeyemi: I do this and find all the prime factors must be consecutative, therefore can only be $2$ or $3$?
– Freeman
Apr 23 '12 at 12:35
1
@LHS think about it. Can $n$ be prime? or Can $n$ be odd?
– Kirthi Raman
Apr 23 '12 at 12:42
@Kv Raman: n can't be prime, unless it equals the totient function, however I'm assuming it can't be odd either?
– Freeman
Apr 23 '12 at 12:47
|
show 1 more comment
Where $phi(n)$ is the Euler phi function, how do you find all $n$ such that $phi(n)|n$?
elementary-number-theory
asked Apr 23 '12 at 12:22
Freeman
88082163
Where $phi(n)$ is the Euler phi function, how do you find all $n$ such that $phi(n)|n$?
elementary-number-theory
elementary-number-theory
asked Apr 23 '12 at 12:22
Freeman
88082163
asked Apr 23 '12 at 12:22
Freeman
88082163
asked Apr 23 '12 at 12:22
Freeman
88082163
asked Apr 23 '12 at 12:22
Freeman
88082163
asked Apr 23 '12 at 12:22
Freeman
88082163
88082163
4
mathforum.org/kb/… and oeis.org/A007694
– Beni Bogosel
Apr 23 '12 at 12:24
2
Use the formula for the totient function in terms of the prime factors of n.
– Dayo Adeyemi
Apr 23 '12 at 12:32
@dayo Adeyemi: I do this and find all the prime factors must be consecutative, therefore can only be $2$ or $3$?
– Freeman
Apr 23 '12 at 12:35
1
@LHS think about it. Can $n$ be prime? or Can $n$ be odd?
– Kirthi Raman
Apr 23 '12 at 12:42
@Kv Raman: n can't be prime, unless it equals the totient function, however I'm assuming it can't be odd either?
– Freeman
Apr 23 '12 at 12:47
|
show 1 more comment
4
mathforum.org/kb/… and oeis.org/A007694
– Beni Bogosel
Apr 23 '12 at 12:24
2
Use the formula for the totient function in terms of the prime factors of n.
– Dayo Adeyemi
Apr 23 '12 at 12:32
@dayo Adeyemi: I do this and find all the prime factors must be consecutative, therefore can only be $2$ or $3$?
– Freeman
Apr 23 '12 at 12:35
1
@LHS think about it. Can $n$ be prime? or Can $n$ be odd?
– Kirthi Raman
Apr 23 '12 at 12:42
@Kv Raman: n can't be prime, unless it equals the totient function, however I'm assuming it can't be odd either?
– Freeman
Apr 23 '12 at 12:47
4
4
mathforum.org/kb/… and oeis.org/A007694
– Beni Bogosel
Apr 23 '12 at 12:24
mathforum.org/kb/… and oeis.org/A007694
– Beni Bogosel
Apr 23 '12 at 12:24
2
2
Use the formula for the totient function in terms of the prime factors of n.
– Dayo Adeyemi
Apr 23 '12 at 12:32
Use the formula for the totient function in terms of the prime factors of n.
– Dayo Adeyemi
Apr 23 '12 at 12:32
@dayo Adeyemi: I do this and find all the prime factors must be consecutative, therefore can only be $2$ or $3$?
– Freeman
Apr 23 '12 at 12:35
@dayo Adeyemi: I do this and find all the prime factors must be consecutative, therefore can only be $2$ or $3$?
– Freeman
Apr 23 '12 at 12:35
1
1
@LHS think about it. Can $n$ be prime? or Can $n$ be odd?
– Kirthi Raman
Apr 23 '12 at 12:42
@LHS think about it. Can $n$ be prime? or Can $n$ be odd?
– Kirthi Raman
Apr 23 '12 at 12:42
@Kv Raman: n can't be prime, unless it equals the totient function, however I'm assuming it can't be odd either?
– Freeman
Apr 23 '12 at 12:47
@Kv Raman: n can't be prime, unless it equals the totient function, however I'm assuming it can't be odd either?
– Freeman
Apr 23 '12 at 12:47
|
show 1 more comment
2 Answers
2
active
oldest
votes
Assume that the prime factorization of $n$ is
$$n = p_1^{a_1} ldots p_k^{a_k}$$
Then the formula for the totient function gives
$$varphi(n) = (p_1 - 1)p_1^{a_1-1}ldots (p_k - 1)p_k^{a_k-1}.$$
If $n>2$, this is always an even number, so $p_1=2$ must appear as a factor. We next observe that $n$ cannot have two odd prime factors. If $a_2>0$ and $a_3>0$, then both $p_2-1$ and $p_3-1$ are even, so $2^{a_1+1}mid varphi(n)$, which is a contradiction.
So $n=2^{a_1}p^{a_2}$ for some prime $p>2$. Here $p-1midvarphi(n)mid n$, so $p-1$
must be a power of two, say $p-1=2^ell$. Then $2^{a_1-1+ell}midvarphi(n)$, so we must have $ell=1$ and $p=3$.
In the end we can verify that $n=2^a3^b$, with $a>0$, $bge0$ is a solution.
This is a very helpful answer, thanks so much! I understand this concept much better now
– Freeman
Apr 23 '12 at 13:29
I feel a bit bad about this. This was meant to address the point left open in m.k.'s answer. But while I was typing, that was deleted. I guess there is a lesson to be learned here?
– Jyrki Lahtonen
Apr 23 '12 at 13:35
Ah. Well I'm very grateful to m.k. As well. Hope they read this!
– Freeman
Apr 23 '12 at 14:03
2
@LHS There are at least a couple prior questions on this topic, so you might find some other prior answers also of interest. Please link them into this question if you find them.
– Bill Dubuque
Apr 23 '12 at 19:40
add a comment |
Quasi-brute-force approach using Maple :
with(numtheory):
for n from 1 to 100 do
if n mod phi(n) = 0 then
print(n);
end if;
end do;
answered Apr 23 '12 at 12:59
Peđa Terzić
7,89022570
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Assume that the prime factorization of $n$ is
$$n = p_1^{a_1} ldots p_k^{a_k}$$
Then the formula for the totient function gives
$$varphi(n) = (p_1 - 1)p_1^{a_1-1}ldots (p_k - 1)p_k^{a_k-1}.$$
If $n>2$, this is always an even number, so $p_1=2$ must appear as a factor. We next observe that $n$ cannot have two odd prime factors. If $a_2>0$ and $a_3>0$, then both $p_2-1$ and $p_3-1$ are even, so $2^{a_1+1}mid varphi(n)$, which is a contradiction.
So $n=2^{a_1}p^{a_2}$ for some prime $p>2$. Here $p-1midvarphi(n)mid n$, so $p-1$
must be a power of two, say $p-1=2^ell$. Then $2^{a_1-1+ell}midvarphi(n)$, so we must have $ell=1$ and $p=3$.
In the end we can verify that $n=2^a3^b$, with $a>0$, $bge0$ is a solution.
This is a very helpful answer, thanks so much! I understand this concept much better now
– Freeman
Apr 23 '12 at 13:29
I feel a bit bad about this. This was meant to address the point left open in m.k.'s answer. But while I was typing, that was deleted. I guess there is a lesson to be learned here?
– Jyrki Lahtonen
Apr 23 '12 at 13:35
Ah. Well I'm very grateful to m.k. As well. Hope they read this!
– Freeman
Apr 23 '12 at 14:03
2
@LHS There are at least a couple prior questions on this topic, so you might find some other prior answers also of interest. Please link them into this question if you find them.
– Bill Dubuque
Apr 23 '12 at 19:40
add a comment |
Assume that the prime factorization of $n$ is
$$n = p_1^{a_1} ldots p_k^{a_k}$$
Then the formula for the totient function gives
$$varphi(n) = (p_1 - 1)p_1^{a_1-1}ldots (p_k - 1)p_k^{a_k-1}.$$
If $n>2$, this is always an even number, so $p_1=2$ must appear as a factor. We next observe that $n$ cannot have two odd prime factors. If $a_2>0$ and $a_3>0$, then both $p_2-1$ and $p_3-1$ are even, so $2^{a_1+1}mid varphi(n)$, which is a contradiction.
So $n=2^{a_1}p^{a_2}$ for some prime $p>2$. Here $p-1midvarphi(n)mid n$, so $p-1$
must be a power of two, say $p-1=2^ell$. Then $2^{a_1-1+ell}midvarphi(n)$, so we must have $ell=1$ and $p=3$.
In the end we can verify that $n=2^a3^b$, with $a>0$, $bge0$ is a solution.
This is a very helpful answer, thanks so much! I understand this concept much better now
– Freeman
Apr 23 '12 at 13:29
I feel a bit bad about this. This was meant to address the point left open in m.k.'s answer. But while I was typing, that was deleted. I guess there is a lesson to be learned here?
– Jyrki Lahtonen
Apr 23 '12 at 13:35
Ah. Well I'm very grateful to m.k. As well. Hope they read this!
– Freeman
Apr 23 '12 at 14:03
2
@LHS There are at least a couple prior questions on this topic, so you might find some other prior answers also of interest. Please link them into this question if you find them.
– Bill Dubuque
Apr 23 '12 at 19:40
add a comment |
Assume that the prime factorization of $n$ is
$$n = p_1^{a_1} ldots p_k^{a_k}$$
Then the formula for the totient function gives
$$varphi(n) = (p_1 - 1)p_1^{a_1-1}ldots (p_k - 1)p_k^{a_k-1}.$$
If $n>2$, this is always an even number, so $p_1=2$ must appear as a factor. We next observe that $n$ cannot have two odd prime factors. If $a_2>0$ and $a_3>0$, then both $p_2-1$ and $p_3-1$ are even, so $2^{a_1+1}mid varphi(n)$, which is a contradiction.
So $n=2^{a_1}p^{a_2}$ for some prime $p>2$. Here $p-1midvarphi(n)mid n$, so $p-1$
must be a power of two, say $p-1=2^ell$. Then $2^{a_1-1+ell}midvarphi(n)$, so we must have $ell=1$ and $p=3$.
In the end we can verify that $n=2^a3^b$, with $a>0$, $bge0$ is a solution.
Assume that the prime factorization of $n$ is
$$n = p_1^{a_1} ldots p_k^{a_k}$$
Then the formula for the totient function gives
$$varphi(n) = (p_1 - 1)p_1^{a_1-1}ldots (p_k - 1)p_k^{a_k-1}.$$
If $n>2$, this is always an even number, so $p_1=2$ must appear as a factor. We next observe that $n$ cannot have two odd prime factors. If $a_2>0$ and $a_3>0$, then both $p_2-1$ and $p_3-1$ are even, so $2^{a_1+1}mid varphi(n)$, which is a contradiction.
So $n=2^{a_1}p^{a_2}$ for some prime $p>2$. Here $p-1midvarphi(n)mid n$, so $p-1$
must be a power of two, say $p-1=2^ell$. Then $2^{a_1-1+ell}midvarphi(n)$, so we must have $ell=1$ and $p=3$.
In the end we can verify that $n=2^a3^b$, with $a>0$, $bge0$ is a solution.
answered Apr 23 '12 at 13:17
community wiki
Jyrki Lahtonen
This is a very helpful answer, thanks so much! I understand this concept much better now
– Freeman
Apr 23 '12 at 13:29
I feel a bit bad about this. This was meant to address the point left open in m.k.'s answer. But while I was typing, that was deleted. I guess there is a lesson to be learned here?
– Jyrki Lahtonen
Apr 23 '12 at 13:35
Ah. Well I'm very grateful to m.k. As well. Hope they read this!
– Freeman
Apr 23 '12 at 14:03
2
@LHS There are at least a couple prior questions on this topic, so you might find some other prior answers also of interest. Please link them into this question if you find them.
– Bill Dubuque
Apr 23 '12 at 19:40
add a comment |
This is a very helpful answer, thanks so much! I understand this concept much better now
– Freeman
Apr 23 '12 at 13:29
I feel a bit bad about this. This was meant to address the point left open in m.k.'s answer. But while I was typing, that was deleted. I guess there is a lesson to be learned here?
– Jyrki Lahtonen
Apr 23 '12 at 13:35
Ah. Well I'm very grateful to m.k. As well. Hope they read this!
– Freeman
Apr 23 '12 at 14:03
2
@LHS There are at least a couple prior questions on this topic, so you might find some other prior answers also of interest. Please link them into this question if you find them.
– Bill Dubuque
Apr 23 '12 at 19:40
This is a very helpful answer, thanks so much! I understand this concept much better now
– Freeman
Apr 23 '12 at 13:29
This is a very helpful answer, thanks so much! I understand this concept much better now
– Freeman
Apr 23 '12 at 13:29
I feel a bit bad about this. This was meant to address the point left open in m.k.'s answer. But while I was typing, that was deleted. I guess there is a lesson to be learned here?
– Jyrki Lahtonen
Apr 23 '12 at 13:35
I feel a bit bad about this. This was meant to address the point left open in m.k.'s answer. But while I was typing, that was deleted. I guess there is a lesson to be learned here?
– Jyrki Lahtonen
Apr 23 '12 at 13:35
Ah. Well I'm very grateful to m.k. As well. Hope they read this!
– Freeman
Apr 23 '12 at 14:03
Ah. Well I'm very grateful to m.k. As well. Hope they read this!
– Freeman
Apr 23 '12 at 14:03
2
2
@LHS There are at least a couple prior questions on this topic, so you might find some other prior answers also of interest. Please link them into this question if you find them.
– Bill Dubuque
Apr 23 '12 at 19:40
@LHS There are at least a couple prior questions on this topic, so you might find some other prior answers also of interest. Please link them into this question if you find them.
– Bill Dubuque
Apr 23 '12 at 19:40
add a comment |
Quasi-brute-force approach using Maple :
with(numtheory):
for n from 1 to 100 do
if n mod phi(n) = 0 then
print(n);
end if;
end do;
answered Apr 23 '12 at 12:59
Peđa Terzić
7,89022570
add a comment |
Quasi-brute-force approach using Maple :
with(numtheory):
for n from 1 to 100 do
if n mod phi(n) = 0 then
print(n);
end if;
end do;
answered Apr 23 '12 at 12:59
Peđa Terzić
7,89022570
add a comment |
Quasi-brute-force approach using Maple :
with(numtheory):
for n from 1 to 100 do
if n mod phi(n) = 0 then
print(n);
end if;
end do;
answered Apr 23 '12 at 12:59
Peđa Terzić
7,89022570
Quasi-brute-force approach using Maple :
with(numtheory):
for n from 1 to 100 do
if n mod phi(n) = 0 then
print(n);
end if;
end do;
answered Apr 23 '12 at 12:59
Peđa Terzić
7,89022570
answered Apr 23 '12 at 12:59
Peđa Terzić
7,89022570
answered Apr 23 '12 at 12:59
Peđa Terzić
7,89022570
answered Apr 23 '12 at 12:59
Peđa Terzić
7,89022570
7,89022570
add a comment |
add a comment |
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4
mathforum.org/kb/… and oeis.org/A007694
– Beni Bogosel
Apr 23 '12 at 12:24
2
Use the formula for the totient function in terms of the prime factors of n.
– Dayo Adeyemi
Apr 23 '12 at 12:32
@dayo Adeyemi: I do this and find all the prime factors must be consecutative, therefore can only be $2$ or $3$?
– Freeman
Apr 23 '12 at 12:35
1
@LHS think about it. Can $n$ be prime? or Can $n$ be odd?
– Kirthi Raman
Apr 23 '12 at 12:42
@Kv Raman: n can't be prime, unless it equals the totient function, however I'm assuming it can't be odd either?
– Freeman
Apr 23 '12 at 12:47