Elementary number theory - perfect numbers [closed]
-1
Let τ(N) and σ(N) the number of divisors and the sum of positive divisors for the positive integer N.
Let k be a positive integer. To solve the system of equations {
τ(a)σ(b)=kb
τ(b)σ(a)=ka
} we can transfer it to the following system: {
τ(a)σ(a)=ka
τ(b)σ(b)=kb
}
If my approach is not correct to this question, could any one advice me what to consider?
elementary-number-theory
edited Jan 4 at 21:57
Matteo Pariset
52
asked Jan 4 at 20:41
PaulPaul
215
closed as off-topic by amWhy, Lee David Chung Lin, jgon, KReiser, onurcanbektas Jan 5 at 3:53
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Lee David Chung Lin, jgon, KReiser, onurcanbektas
If this question can be reworded to fit the rules in the help center, please edit the question.
|
show 3 more comments
-1
Let τ(N) and σ(N) the number of divisors and the sum of positive divisors for the positive integer N.
Let k be a positive integer. To solve the system of equations {
τ(a)σ(b)=kb
τ(b)σ(a)=ka
} we can transfer it to the following system: {
τ(a)σ(a)=ka
τ(b)σ(b)=kb
}
If my approach is not correct to this question, could any one advice me what to consider?
elementary-number-theory
edited Jan 4 at 21:57
Matteo Pariset
52
asked Jan 4 at 20:41
PaulPaul
215
closed as off-topic by amWhy, Lee David Chung Lin, jgon, KReiser, onurcanbektas Jan 5 at 3:53
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Lee David Chung Lin, jgon, KReiser, onurcanbektas
If this question can be reworded to fit the rules in the help center, please edit the question.
1
I don't know how to solve these systems. But your approach is definitely not correct -- because the two systems are different, not equivalent to each other. So you can't "transfer" one into the other. They are only equivalent under the additional assumption that $tau(a)=tau(b)$.
– zipirovich
Jan 4 at 22:16
my approach: If we multiply both equations in the initial system we get τ(a)σ(b)τ(a)σ(a)=kbka or equivalently: ((σ(a)τ(a))/k)((σ(b)τ(b))/k)=ab One possible way is to let: ((σ(a)τ(a))/k)=a ((σ(b)τ(b))/k)=b which then implies: σ(a)τ(a)=ka σ(b)τ(b)=kb
– Paul
Jan 5 at 3:41
"One possible way" does NOT mean that these have to be the values. What if another possible way holds true? For example, assume we know that $xy=6$. One possible way is to let $x=2$ and $y=3$. Does it mean that $x$ is definitely $2$ and $y$ is definitely $3$? Obviously, not at all! It might as well be that $x=1$ and $y=6$ or any of the other possibilities.
– zipirovich
Jan 5 at 4:02
Yes you are right. Thank you for you kind comment. What if we take a specific value for $k$, for instance $k=3$?
– Paul
Jan 5 at 4:19
What do perfect numbers have to do with the given equations ?
– Peter
Jan 5 at 10:58
|
show 3 more comments
-1
-1
-1
Let τ(N) and σ(N) the number of divisors and the sum of positive divisors for the positive integer N.
Let k be a positive integer. To solve the system of equations {
τ(a)σ(b)=kb
τ(b)σ(a)=ka
} we can transfer it to the following system: {
τ(a)σ(a)=ka
τ(b)σ(b)=kb
}
If my approach is not correct to this question, could any one advice me what to consider?
elementary-number-theory
edited Jan 4 at 21:57
Matteo Pariset
52
asked Jan 4 at 20:41
PaulPaul
215
Let τ(N) and σ(N) the number of divisors and the sum of positive divisors for the positive integer N.
Let k be a positive integer. To solve the system of equations {
τ(a)σ(b)=kb
τ(b)σ(a)=ka
} we can transfer it to the following system: {
τ(a)σ(a)=ka
τ(b)σ(b)=kb
}
If my approach is not correct to this question, could any one advice me what to consider?
elementary-number-theory
elementary-number-theory
edited Jan 4 at 21:57
Matteo Pariset
52
asked Jan 4 at 20:41
PaulPaul
215
edited Jan 4 at 21:57
Matteo Pariset
52
asked Jan 4 at 20:41
PaulPaul
215
edited Jan 4 at 21:57
Matteo Pariset
52
edited Jan 4 at 21:57
Matteo Pariset
52
edited Jan 4 at 21:57
Matteo Pariset
52
52
asked Jan 4 at 20:41
PaulPaul
215
asked Jan 4 at 20:41
PaulPaul
215
asked Jan 4 at 20:41
PaulPaul
215
215
closed as off-topic by amWhy, Lee David Chung Lin, jgon, KReiser, onurcanbektas Jan 5 at 3:53
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Lee David Chung Lin, jgon, KReiser, onurcanbektas
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by amWhy, Lee David Chung Lin, jgon, KReiser, onurcanbektas Jan 5 at 3:53
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Lee David Chung Lin, jgon, KReiser, onurcanbektas
If this question can be reworded to fit the rules in the help center, please edit the question.
1
I don't know how to solve these systems. But your approach is definitely not correct -- because the two systems are different, not equivalent to each other. So you can't "transfer" one into the other. They are only equivalent under the additional assumption that $tau(a)=tau(b)$.
– zipirovich
Jan 4 at 22:16
my approach: If we multiply both equations in the initial system we get τ(a)σ(b)τ(a)σ(a)=kbka or equivalently: ((σ(a)τ(a))/k)((σ(b)τ(b))/k)=ab One possible way is to let: ((σ(a)τ(a))/k)=a ((σ(b)τ(b))/k)=b which then implies: σ(a)τ(a)=ka σ(b)τ(b)=kb
– Paul
Jan 5 at 3:41
"One possible way" does NOT mean that these have to be the values. What if another possible way holds true? For example, assume we know that $xy=6$. One possible way is to let $x=2$ and $y=3$. Does it mean that $x$ is definitely $2$ and $y$ is definitely $3$? Obviously, not at all! It might as well be that $x=1$ and $y=6$ or any of the other possibilities.
– zipirovich
Jan 5 at 4:02
Yes you are right. Thank you for you kind comment. What if we take a specific value for $k$, for instance $k=3$?
– Paul
Jan 5 at 4:19
What do perfect numbers have to do with the given equations ?
– Peter
Jan 5 at 10:58
|
show 3 more comments
1
I don't know how to solve these systems. But your approach is definitely not correct -- because the two systems are different, not equivalent to each other. So you can't "transfer" one into the other. They are only equivalent under the additional assumption that $tau(a)=tau(b)$.
– zipirovich
Jan 4 at 22:16
my approach: If we multiply both equations in the initial system we get τ(a)σ(b)τ(a)σ(a)=kbka or equivalently: ((σ(a)τ(a))/k)((σ(b)τ(b))/k)=ab One possible way is to let: ((σ(a)τ(a))/k)=a ((σ(b)τ(b))/k)=b which then implies: σ(a)τ(a)=ka σ(b)τ(b)=kb
– Paul
Jan 5 at 3:41
"One possible way" does NOT mean that these have to be the values. What if another possible way holds true? For example, assume we know that $xy=6$. One possible way is to let $x=2$ and $y=3$. Does it mean that $x$ is definitely $2$ and $y$ is definitely $3$? Obviously, not at all! It might as well be that $x=1$ and $y=6$ or any of the other possibilities.
– zipirovich
Jan 5 at 4:02
Yes you are right. Thank you for you kind comment. What if we take a specific value for $k$, for instance $k=3$?
– Paul
Jan 5 at 4:19
What do perfect numbers have to do with the given equations ?
– Peter
Jan 5 at 10:58
1
1
I don't know how to solve these systems. But your approach is definitely not correct -- because the two systems are different, not equivalent to each other. So you can't "transfer" one into the other. They are only equivalent under the additional assumption that $tau(a)=tau(b)$.
– zipirovich
Jan 4 at 22:16
I don't know how to solve these systems. But your approach is definitely not correct -- because the two systems are different, not equivalent to each other. So you can't "transfer" one into the other. They are only equivalent under the additional assumption that $tau(a)=tau(b)$.
– zipirovich
Jan 4 at 22:16
my approach: If we multiply both equations in the initial system we get τ(a)σ(b)τ(a)σ(a)=kbka or equivalently: ((σ(a)τ(a))/k)((σ(b)τ(b))/k)=ab One possible way is to let: ((σ(a)τ(a))/k)=a ((σ(b)τ(b))/k)=b which then implies: σ(a)τ(a)=ka σ(b)τ(b)=kb
– Paul
Jan 5 at 3:41
my approach: If we multiply both equations in the initial system we get τ(a)σ(b)τ(a)σ(a)=kbka or equivalently: ((σ(a)τ(a))/k)((σ(b)τ(b))/k)=ab One possible way is to let: ((σ(a)τ(a))/k)=a ((σ(b)τ(b))/k)=b which then implies: σ(a)τ(a)=ka σ(b)τ(b)=kb
– Paul
Jan 5 at 3:41
"One possible way" does NOT mean that these have to be the values. What if another possible way holds true? For example, assume we know that $xy=6$. One possible way is to let $x=2$ and $y=3$. Does it mean that $x$ is definitely $2$ and $y$ is definitely $3$? Obviously, not at all! It might as well be that $x=1$ and $y=6$ or any of the other possibilities.
– zipirovich
Jan 5 at 4:02
"One possible way" does NOT mean that these have to be the values. What if another possible way holds true? For example, assume we know that $xy=6$. One possible way is to let $x=2$ and $y=3$. Does it mean that $x$ is definitely $2$ and $y$ is definitely $3$? Obviously, not at all! It might as well be that $x=1$ and $y=6$ or any of the other possibilities.
– zipirovich
Jan 5 at 4:02
Yes you are right. Thank you for you kind comment. What if we take a specific value for $k$, for instance $k=3$?
– Paul
Jan 5 at 4:19
Yes you are right. Thank you for you kind comment. What if we take a specific value for $k$, for instance $k=3$?
– Paul
Jan 5 at 4:19
What do perfect numbers have to do with the given equations ?
– Peter
Jan 5 at 10:58
What do perfect numbers have to do with the given equations ?
– Peter
Jan 5 at 10:58
|
show 3 more comments
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I don't know how to solve these systems. But your approach is definitely not correct -- because the two systems are different, not equivalent to each other. So you can't "transfer" one into the other. They are only equivalent under the additional assumption that $tau(a)=tau(b)$.
– zipirovich
Jan 4 at 22:16
my approach: If we multiply both equations in the initial system we get τ(a)σ(b)τ(a)σ(a)=kbka or equivalently: ((σ(a)τ(a))/k)((σ(b)τ(b))/k)=ab One possible way is to let: ((σ(a)τ(a))/k)=a ((σ(b)τ(b))/k)=b which then implies: σ(a)τ(a)=ka σ(b)τ(b)=kb
– Paul
Jan 5 at 3:41
"One possible way" does NOT mean that these have to be the values. What if another possible way holds true? For example, assume we know that $xy=6$. One possible way is to let $x=2$ and $y=3$. Does it mean that $x$ is definitely $2$ and $y$ is definitely $3$? Obviously, not at all! It might as well be that $x=1$ and $y=6$ or any of the other possibilities.
– zipirovich
Jan 5 at 4:02
Yes you are right. Thank you for you kind comment. What if we take a specific value for $k$, for instance $k=3$?
– Paul
Jan 5 at 4:19
What do perfect numbers have to do with the given equations ?
– Peter
Jan 5 at 10:58