Find all values of $n$ such that $varphi(n) = n/6$. [duplicate]
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This question already has an answer here:
Is there $phi(n)=n/6$
2 answers
Using the product formula (the formula with the prime factors of $n$), I got
$$1=6frac{(P_1-1)}{P_1}frac{(P_2-1)}{P_2}cdotsfrac{(P_k-1)}{P_k},.$$
number-theory elementary-number-theory prime-numbers diophantine-equations totient-function
edited Dec 12 '18 at 3:28
Batominovski
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asked Dec 12 '18 at 0:14
MeganMegan
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marked as duplicate by Shailesh, Leucippus, Brahadeesh, Dietrich Burde number-theory
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Dec 12 '18 at 9:32
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Is there $phi(n)=n/6$
2 answers
Using the product formula (the formula with the prime factors of $n$), I got
$$1=6frac{(P_1-1)}{P_1}frac{(P_2-1)}{P_2}cdotsfrac{(P_k-1)}{P_k},.$$
number-theory elementary-number-theory prime-numbers diophantine-equations totient-function
edited Dec 12 '18 at 3:28
Batominovski
1
asked Dec 12 '18 at 0:14
MeganMegan
32
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marked as duplicate by Shailesh, Leucippus, Brahadeesh, Dietrich Burde number-theory
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Dec 12 '18 at 9:32
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
2
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Welcome to MSE! For some basic information about writing mathematics at this site see, e.g., here, here, here and here. Additionally, if you are looking for prsomebody to check your proof, please update your post to include the tag proof-verification.
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– platty
Dec 12 '18 at 0:19
add a comment |
$begingroup$
This question already has an answer here:
Is there $phi(n)=n/6$
2 answers
Using the product formula (the formula with the prime factors of $n$), I got
$$1=6frac{(P_1-1)}{P_1}frac{(P_2-1)}{P_2}cdotsfrac{(P_k-1)}{P_k},.$$
number-theory elementary-number-theory prime-numbers diophantine-equations totient-function
edited Dec 12 '18 at 3:28
Batominovski
1
asked Dec 12 '18 at 0:14
MeganMegan
32
$endgroup$
This question already has an answer here:
Is there $phi(n)=n/6$
2 answers
Using the product formula (the formula with the prime factors of $n$), I got
$$1=6frac{(P_1-1)}{P_1}frac{(P_2-1)}{P_2}cdotsfrac{(P_k-1)}{P_k},.$$
This question already has an answer here:
Is there $phi(n)=n/6$
2 answers
number-theory elementary-number-theory prime-numbers diophantine-equations totient-function
number-theory elementary-number-theory prime-numbers diophantine-equations totient-function
edited Dec 12 '18 at 3:28
Batominovski
1
asked Dec 12 '18 at 0:14
MeganMegan
32
edited Dec 12 '18 at 3:28
Batominovski
1
asked Dec 12 '18 at 0:14
MeganMegan
32
edited Dec 12 '18 at 3:28
Batominovski
1
edited Dec 12 '18 at 3:28
Batominovski
1
edited Dec 12 '18 at 3:28
Batominovski
1
1
asked Dec 12 '18 at 0:14
MeganMegan
32
asked Dec 12 '18 at 0:14
MeganMegan
32
asked Dec 12 '18 at 0:14
MeganMegan
32
32
marked as duplicate by Shailesh, Leucippus, Brahadeesh, Dietrich Burde number-theory
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Dec 12 '18 at 9:32
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Dec 12 '18 at 9:32
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
2
$begingroup$
Welcome to MSE! For some basic information about writing mathematics at this site see, e.g., here, here, here and here. Additionally, if you are looking for prsomebody to check your proof, please update your post to include the tag proof-verification.
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– platty
Dec 12 '18 at 0:19
add a comment |
2
$begingroup$
Welcome to MSE! For some basic information about writing mathematics at this site see, e.g., here, here, here and here. Additionally, if you are looking for prsomebody to check your proof, please update your post to include the tag proof-verification.
$endgroup$
– platty
Dec 12 '18 at 0:19
2
2
$begingroup$
Welcome to MSE! For some basic information about writing mathematics at this site see, e.g., here, here, here and here. Additionally, if you are looking for prsomebody to check your proof, please update your post to include the tag proof-verification.
$endgroup$
– platty
Dec 12 '18 at 0:19
$begingroup$
Welcome to MSE! For some basic information about writing mathematics at this site see, e.g., here, here, here and here. Additionally, if you are looking for prsomebody to check your proof, please update your post to include the tag proof-verification.
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– platty
Dec 12 '18 at 0:19
add a comment |
2 Answers
2
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oldest
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i don't think the proof is enough, as it stands. There is a small ingredient. We are taking the product, with one or more primes, of $frac{p-1}{p}.$
There are two types of potential problems. Some cancellation is possible, putting the product in lowest terms may be a bit surprising. Second, we can get the product arbitrarily close to $frac{1}{6}$ byt choosing many large primes.
However: the collection of chosen primes is finite, and one of them is largest, call that $q.$ For any of the other, smaller primes, since $p<q,$ we also have $p-1 < q.$ That means that the numerator of the product fraction is not divisible by $q.$ When we put the fraction in lowest terms, the denominator is still divisible by $q.$
For your problem, this means that the largest prime involved, $q,$ must be a factor of $6.$ The possibilities for the product are just:
$$ { }, frac{1}{1} $$
$$ { 2}, frac{1}{2} $$
$$ { 3}, frac{2}{3} $$
$$ { 2,3}, frac{1}{3} $$
answered Dec 12 '18 at 0:54
Will JagyWill Jagy
103k5101200
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add a comment |
$begingroup$
Hints:
We know that if $;p_1,...,p_k;$ are the different primes diving $;n;$ , then
$$varphi(n)=nprod_{j=1}^kleft(1-frac1{p_j}right)$$
Well, take it now...And I couldn't completely understand what you wrote, but I think you're on the right track: there is no solution.
answered Dec 12 '18 at 0:23
DonAntonioDonAntonio
178k1493229
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$begingroup$
i used the same definition but because i couldnt type the product symbol i used the definition of the product....anyway thanks for verifying that there is no solution
$endgroup$
– Megan
Dec 12 '18 at 0:27
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
i don't think the proof is enough, as it stands. There is a small ingredient. We are taking the product, with one or more primes, of $frac{p-1}{p}.$
There are two types of potential problems. Some cancellation is possible, putting the product in lowest terms may be a bit surprising. Second, we can get the product arbitrarily close to $frac{1}{6}$ byt choosing many large primes.
However: the collection of chosen primes is finite, and one of them is largest, call that $q.$ For any of the other, smaller primes, since $p<q,$ we also have $p-1 < q.$ That means that the numerator of the product fraction is not divisible by $q.$ When we put the fraction in lowest terms, the denominator is still divisible by $q.$
For your problem, this means that the largest prime involved, $q,$ must be a factor of $6.$ The possibilities for the product are just:
$$ { }, frac{1}{1} $$
$$ { 2}, frac{1}{2} $$
$$ { 3}, frac{2}{3} $$
$$ { 2,3}, frac{1}{3} $$
answered Dec 12 '18 at 0:54
Will JagyWill Jagy
103k5101200
$endgroup$
add a comment |
$begingroup$
i don't think the proof is enough, as it stands. There is a small ingredient. We are taking the product, with one or more primes, of $frac{p-1}{p}.$
There are two types of potential problems. Some cancellation is possible, putting the product in lowest terms may be a bit surprising. Second, we can get the product arbitrarily close to $frac{1}{6}$ byt choosing many large primes.
However: the collection of chosen primes is finite, and one of them is largest, call that $q.$ For any of the other, smaller primes, since $p<q,$ we also have $p-1 < q.$ That means that the numerator of the product fraction is not divisible by $q.$ When we put the fraction in lowest terms, the denominator is still divisible by $q.$
For your problem, this means that the largest prime involved, $q,$ must be a factor of $6.$ The possibilities for the product are just:
$$ { }, frac{1}{1} $$
$$ { 2}, frac{1}{2} $$
$$ { 3}, frac{2}{3} $$
$$ { 2,3}, frac{1}{3} $$
answered Dec 12 '18 at 0:54
Will JagyWill Jagy
103k5101200
$endgroup$
add a comment |
$begingroup$
i don't think the proof is enough, as it stands. There is a small ingredient. We are taking the product, with one or more primes, of $frac{p-1}{p}.$
There are two types of potential problems. Some cancellation is possible, putting the product in lowest terms may be a bit surprising. Second, we can get the product arbitrarily close to $frac{1}{6}$ byt choosing many large primes.
However: the collection of chosen primes is finite, and one of them is largest, call that $q.$ For any of the other, smaller primes, since $p<q,$ we also have $p-1 < q.$ That means that the numerator of the product fraction is not divisible by $q.$ When we put the fraction in lowest terms, the denominator is still divisible by $q.$
For your problem, this means that the largest prime involved, $q,$ must be a factor of $6.$ The possibilities for the product are just:
$$ { }, frac{1}{1} $$
$$ { 2}, frac{1}{2} $$
$$ { 3}, frac{2}{3} $$
$$ { 2,3}, frac{1}{3} $$
answered Dec 12 '18 at 0:54
Will JagyWill Jagy
103k5101200
$endgroup$
i don't think the proof is enough, as it stands. There is a small ingredient. We are taking the product, with one or more primes, of $frac{p-1}{p}.$
There are two types of potential problems. Some cancellation is possible, putting the product in lowest terms may be a bit surprising. Second, we can get the product arbitrarily close to $frac{1}{6}$ byt choosing many large primes.
However: the collection of chosen primes is finite, and one of them is largest, call that $q.$ For any of the other, smaller primes, since $p<q,$ we also have $p-1 < q.$ That means that the numerator of the product fraction is not divisible by $q.$ When we put the fraction in lowest terms, the denominator is still divisible by $q.$
For your problem, this means that the largest prime involved, $q,$ must be a factor of $6.$ The possibilities for the product are just:
$$ { }, frac{1}{1} $$
$$ { 2}, frac{1}{2} $$
$$ { 3}, frac{2}{3} $$
$$ { 2,3}, frac{1}{3} $$
answered Dec 12 '18 at 0:54
Will JagyWill Jagy
103k5101200
answered Dec 12 '18 at 0:54
Will JagyWill Jagy
103k5101200
answered Dec 12 '18 at 0:54
Will JagyWill Jagy
103k5101200
answered Dec 12 '18 at 0:54
Will JagyWill Jagy
103k5101200
103k5101200
add a comment |
add a comment |
$begingroup$
Hints:
We know that if $;p_1,...,p_k;$ are the different primes diving $;n;$ , then
$$varphi(n)=nprod_{j=1}^kleft(1-frac1{p_j}right)$$
Well, take it now...And I couldn't completely understand what you wrote, but I think you're on the right track: there is no solution.
answered Dec 12 '18 at 0:23
DonAntonioDonAntonio
178k1493229
$endgroup$
$begingroup$
i used the same definition but because i couldnt type the product symbol i used the definition of the product....anyway thanks for verifying that there is no solution
$endgroup$
– Megan
Dec 12 '18 at 0:27
add a comment |
$begingroup$
Hints:
We know that if $;p_1,...,p_k;$ are the different primes diving $;n;$ , then
$$varphi(n)=nprod_{j=1}^kleft(1-frac1{p_j}right)$$
Well, take it now...And I couldn't completely understand what you wrote, but I think you're on the right track: there is no solution.
answered Dec 12 '18 at 0:23
DonAntonioDonAntonio
178k1493229
$endgroup$
$begingroup$
i used the same definition but because i couldnt type the product symbol i used the definition of the product....anyway thanks for verifying that there is no solution
$endgroup$
– Megan
Dec 12 '18 at 0:27
add a comment |
$begingroup$
Hints:
We know that if $;p_1,...,p_k;$ are the different primes diving $;n;$ , then
$$varphi(n)=nprod_{j=1}^kleft(1-frac1{p_j}right)$$
Well, take it now...And I couldn't completely understand what you wrote, but I think you're on the right track: there is no solution.
answered Dec 12 '18 at 0:23
DonAntonioDonAntonio
178k1493229
$endgroup$
Hints:
We know that if $;p_1,...,p_k;$ are the different primes diving $;n;$ , then
$$varphi(n)=nprod_{j=1}^kleft(1-frac1{p_j}right)$$
Well, take it now...And I couldn't completely understand what you wrote, but I think you're on the right track: there is no solution.
answered Dec 12 '18 at 0:23
DonAntonioDonAntonio
178k1493229
answered Dec 12 '18 at 0:23
DonAntonioDonAntonio
178k1493229
answered Dec 12 '18 at 0:23
DonAntonioDonAntonio
178k1493229
answered Dec 12 '18 at 0:23
DonAntonioDonAntonio
178k1493229
178k1493229
$begingroup$
i used the same definition but because i couldnt type the product symbol i used the definition of the product....anyway thanks for verifying that there is no solution
$endgroup$
– Megan
Dec 12 '18 at 0:27
add a comment |
$begingroup$
i used the same definition but because i couldnt type the product symbol i used the definition of the product....anyway thanks for verifying that there is no solution
$endgroup$
– Megan
Dec 12 '18 at 0:27
$begingroup$
i used the same definition but because i couldnt type the product symbol i used the definition of the product....anyway thanks for verifying that there is no solution
$endgroup$
– Megan
Dec 12 '18 at 0:27
$begingroup$
i used the same definition but because i couldnt type the product symbol i used the definition of the product....anyway thanks for verifying that there is no solution
$endgroup$
– Megan
Dec 12 '18 at 0:27
add a comment |
2
$begingroup$
Welcome to MSE! For some basic information about writing mathematics at this site see, e.g., here, here, here and here. Additionally, if you are looking for prsomebody to check your proof, please update your post to include the tag proof-verification.
$endgroup$
– platty
Dec 12 '18 at 0:19