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Do there exist distinct integers $p$ and $q$ such that $dfracpn = dfracqn+1$ for some integer $n$? [closed]
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Do there exist distinct integers $p$ and $q$ such that $dfracpn = dfracqn+1$ for some integer $n$, given $p$ is less than $n$ and likewise $q$ is less than $n+1$?
Assume that $dfracpn$ and $dfracqn+1$ can be rational, non-integers.
edit: changed less than inequality constraint to strictly less than.
number-theory elementary-number-theory
asked Jul 14 at 23:53
user115660
63
closed as off-topic by Adrian Keister, amWhy, Shailesh, Alex Francisco, Parcly Taxel Jul 15 at 3:55
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Adrian Keister, amWhy, Shailesh, Alex Francisco, Parcly Taxel
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up vote
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Do there exist distinct integers $p$ and $q$ such that $dfracpn = dfracqn+1$ for some integer $n$, given $p$ is less than $n$ and likewise $q$ is less than $n+1$?
Assume that $dfracpn$ and $dfracqn+1$ can be rational, non-integers.
edit: changed less than inequality constraint to strictly less than.
number-theory elementary-number-theory
asked Jul 14 at 23:53
user115660
63
closed as off-topic by Adrian Keister, amWhy, Shailesh, Alex Francisco, Parcly Taxel Jul 15 at 3:55
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Adrian Keister, amWhy, Shailesh, Alex Francisco, Parcly Taxel
p and q must be distinct.
– user115660
Jul 14 at 23:55
2
The question isn't clear. Why not take $p=n,q=n+1$?
– lulu
Jul 14 at 23:56
1
P=n and q=n+1 works
– mm8511
Jul 14 at 23:57
Mmm, I'll update it and remove the equality constraint.
– user115660
Jul 14 at 23:58
Since $gcd(n,n+1)=1$ it is clear that $n$ divides $p$ and $n-1$ divides $q$ (assuming that $p,q>0$).
– lulu
Jul 15 at 0:01
 |Â
show 1 more comment
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Do there exist distinct integers $p$ and $q$ such that $dfracpn = dfracqn+1$ for some integer $n$, given $p$ is less than $n$ and likewise $q$ is less than $n+1$?
Assume that $dfracpn$ and $dfracqn+1$ can be rational, non-integers.
edit: changed less than inequality constraint to strictly less than.
number-theory elementary-number-theory
asked Jul 14 at 23:53
user115660
63
Do there exist distinct integers $p$ and $q$ such that $dfracpn = dfracqn+1$ for some integer $n$, given $p$ is less than $n$ and likewise $q$ is less than $n+1$?
Assume that $dfracpn$ and $dfracqn+1$ can be rational, non-integers.
edit: changed less than inequality constraint to strictly less than.
number-theory elementary-number-theory
asked Jul 14 at 23:53
user115660
63
edited Jul 15 at 0:12
asked Jul 14 at 23:53
user115660
63
asked Jul 14 at 23:53
user115660
63
asked Jul 14 at 23:53
user115660
63
63
closed as off-topic by Adrian Keister, amWhy, Shailesh, Alex Francisco, Parcly Taxel Jul 15 at 3:55
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Adrian Keister, amWhy, Shailesh, Alex Francisco, Parcly Taxel
closed as off-topic by Adrian Keister, amWhy, Shailesh, Alex Francisco, Parcly Taxel Jul 15 at 3:55
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Adrian Keister, amWhy, Shailesh, Alex Francisco, Parcly Taxel
p and q must be distinct.
– user115660
Jul 14 at 23:55
2
The question isn't clear. Why not take $p=n,q=n+1$?
– lulu
Jul 14 at 23:56
1
P=n and q=n+1 works
– mm8511
Jul 14 at 23:57
Mmm, I'll update it and remove the equality constraint.
– user115660
Jul 14 at 23:58
Since $gcd(n,n+1)=1$ it is clear that $n$ divides $p$ and $n-1$ divides $q$ (assuming that $p,q>0$).
– lulu
Jul 15 at 0:01
 |Â
show 1 more comment
p and q must be distinct.
– user115660
Jul 14 at 23:55
2
The question isn't clear. Why not take $p=n,q=n+1$?
– lulu
Jul 14 at 23:56
1
P=n and q=n+1 works
– mm8511
Jul 14 at 23:57
Mmm, I'll update it and remove the equality constraint.
– user115660
Jul 14 at 23:58
Since $gcd(n,n+1)=1$ it is clear that $n$ divides $p$ and $n-1$ divides $q$ (assuming that $p,q>0$).
– lulu
Jul 15 at 0:01
p and q must be distinct.
– user115660
Jul 14 at 23:55
p and q must be distinct.
– user115660
Jul 14 at 23:55
2
2
The question isn't clear. Why not take $p=n,q=n+1$?
– lulu
Jul 14 at 23:56
The question isn't clear. Why not take $p=n,q=n+1$?
– lulu
Jul 14 at 23:56
1
1
P=n and q=n+1 works
– mm8511
Jul 14 at 23:57
P=n and q=n+1 works
– mm8511
Jul 14 at 23:57
Mmm, I'll update it and remove the equality constraint.
– user115660
Jul 14 at 23:58
Mmm, I'll update it and remove the equality constraint.
– user115660
Jul 14 at 23:58
Since $gcd(n,n+1)=1$ it is clear that $n$ divides $p$ and $n-1$ divides $q$ (assuming that $p,q>0$).
– lulu
Jul 15 at 0:01
Since $gcd(n,n+1)=1$ it is clear that $n$ divides $p$ and $n-1$ divides $q$ (assuming that $p,q>0$).
– lulu
Jul 15 at 0:01
 |Â
show 1 more comment
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
Your equation can be rewritten as $frac nn+1=frac pq$, therefore the answer is yes, since $n$ and $n+1$ are necessarily coprime, the fraction $frac nn+1$ is reduced (and the unique integers are the obvious $p=n$, $q=n+1$).
Edit: A direct consequence of the above is that the answer to the new question is no. Since the only (unique!) solution is to have $p$ and $q$ respectively equal to $n$ and $n+1$.
answered Jul 14 at 23:56
Arnaud Mortier
19.2k22159
This is correct! I've updated my question so that the integers must be distinct, not necessarily unique, and constraint is now strictly less than.
– user115660
Jul 15 at 0:02
@user115660 Edited.
– Arnaud Mortier
Jul 15 at 0:06
Ah ha, interesting. By asking the wrong question, the provided answer could be used to infer the solution to the intended question. Thank you @Arnaud Mortier.
– user115660
Jul 15 at 0:10
@user115660 You're welcome.
– Arnaud Mortier
Jul 15 at 0:10
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Your equation can be rewritten as $frac nn+1=frac pq$, therefore the answer is yes, since $n$ and $n+1$ are necessarily coprime, the fraction $frac nn+1$ is reduced (and the unique integers are the obvious $p=n$, $q=n+1$).
Edit: A direct consequence of the above is that the answer to the new question is no. Since the only (unique!) solution is to have $p$ and $q$ respectively equal to $n$ and $n+1$.
answered Jul 14 at 23:56
Arnaud Mortier
19.2k22159
This is correct! I've updated my question so that the integers must be distinct, not necessarily unique, and constraint is now strictly less than.
– user115660
Jul 15 at 0:02
@user115660 Edited.
– Arnaud Mortier
Jul 15 at 0:06
Ah ha, interesting. By asking the wrong question, the provided answer could be used to infer the solution to the intended question. Thank you @Arnaud Mortier.
– user115660
Jul 15 at 0:10
@user115660 You're welcome.
– Arnaud Mortier
Jul 15 at 0:10
add a comment |Â
up vote
2
down vote
accepted
Your equation can be rewritten as $frac nn+1=frac pq$, therefore the answer is yes, since $n$ and $n+1$ are necessarily coprime, the fraction $frac nn+1$ is reduced (and the unique integers are the obvious $p=n$, $q=n+1$).
Edit: A direct consequence of the above is that the answer to the new question is no. Since the only (unique!) solution is to have $p$ and $q$ respectively equal to $n$ and $n+1$.
answered Jul 14 at 23:56
Arnaud Mortier
19.2k22159
This is correct! I've updated my question so that the integers must be distinct, not necessarily unique, and constraint is now strictly less than.
– user115660
Jul 15 at 0:02
@user115660 Edited.
– Arnaud Mortier
Jul 15 at 0:06
Ah ha, interesting. By asking the wrong question, the provided answer could be used to infer the solution to the intended question. Thank you @Arnaud Mortier.
– user115660
Jul 15 at 0:10
@user115660 You're welcome.
– Arnaud Mortier
Jul 15 at 0:10
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Your equation can be rewritten as $frac nn+1=frac pq$, therefore the answer is yes, since $n$ and $n+1$ are necessarily coprime, the fraction $frac nn+1$ is reduced (and the unique integers are the obvious $p=n$, $q=n+1$).
Edit: A direct consequence of the above is that the answer to the new question is no. Since the only (unique!) solution is to have $p$ and $q$ respectively equal to $n$ and $n+1$.
answered Jul 14 at 23:56
Arnaud Mortier
19.2k22159
Your equation can be rewritten as $frac nn+1=frac pq$, therefore the answer is yes, since $n$ and $n+1$ are necessarily coprime, the fraction $frac nn+1$ is reduced (and the unique integers are the obvious $p=n$, $q=n+1$).
Edit: A direct consequence of the above is that the answer to the new question is no. Since the only (unique!) solution is to have $p$ and $q$ respectively equal to $n$ and $n+1$.
answered Jul 14 at 23:56
Arnaud Mortier
19.2k22159
edited Jul 15 at 0:05
answered Jul 14 at 23:56
Arnaud Mortier
19.2k22159
answered Jul 14 at 23:56
Arnaud Mortier
19.2k22159
answered Jul 14 at 23:56
Arnaud Mortier
19.2k22159
19.2k22159
This is correct! I've updated my question so that the integers must be distinct, not necessarily unique, and constraint is now strictly less than.
– user115660
Jul 15 at 0:02
@user115660 Edited.
– Arnaud Mortier
Jul 15 at 0:06
Ah ha, interesting. By asking the wrong question, the provided answer could be used to infer the solution to the intended question. Thank you @Arnaud Mortier.
– user115660
Jul 15 at 0:10
@user115660 You're welcome.
– Arnaud Mortier
Jul 15 at 0:10
add a comment |Â
This is correct! I've updated my question so that the integers must be distinct, not necessarily unique, and constraint is now strictly less than.
– user115660
Jul 15 at 0:02
@user115660 Edited.
– Arnaud Mortier
Jul 15 at 0:06
Ah ha, interesting. By asking the wrong question, the provided answer could be used to infer the solution to the intended question. Thank you @Arnaud Mortier.
– user115660
Jul 15 at 0:10
@user115660 You're welcome.
– Arnaud Mortier
Jul 15 at 0:10
This is correct! I've updated my question so that the integers must be distinct, not necessarily unique, and constraint is now strictly less than.
– user115660
Jul 15 at 0:02
This is correct! I've updated my question so that the integers must be distinct, not necessarily unique, and constraint is now strictly less than.
– user115660
Jul 15 at 0:02
@user115660 Edited.
– Arnaud Mortier
Jul 15 at 0:06
@user115660 Edited.
– Arnaud Mortier
Jul 15 at 0:06
Ah ha, interesting. By asking the wrong question, the provided answer could be used to infer the solution to the intended question. Thank you @Arnaud Mortier.
– user115660
Jul 15 at 0:10
Ah ha, interesting. By asking the wrong question, the provided answer could be used to infer the solution to the intended question. Thank you @Arnaud Mortier.
– user115660
Jul 15 at 0:10
@user115660 You're welcome.
– Arnaud Mortier
Jul 15 at 0:10
@user115660 You're welcome.
– Arnaud Mortier
Jul 15 at 0:10
add a comment |Â
p and q must be distinct.
– user115660
Jul 14 at 23:55
2
The question isn't clear. Why not take $p=n,q=n+1$?
– lulu
Jul 14 at 23:56
1
P=n and q=n+1 works
– mm8511
Jul 14 at 23:57
Mmm, I'll update it and remove the equality constraint.
– user115660
Jul 14 at 23:58
Since $gcd(n,n+1)=1$ it is clear that $n$ divides $p$ and $n-1$ divides $q$ (assuming that $p,q>0$).
– lulu
Jul 15 at 0:01