$m^{5}n+7$ and $n^{5}m+7$ are cubes of integer. [on hold]
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Find integer numbers $m$, $n$ such that $m^{5}n+7$ and $n^{5}m+7$ are cubes of integer.
I guess the only pair is $(1;1)$, but I don't know what way we should follow to solve this problem. Help me please.
number-theory elementary-number-theory diophantine-equations problem-solving
edited Nov 14 at 15:03
Batominovski
31.5k23187
asked Nov 14 at 14:22
Trong Tuan
585
put on hold as off-topic by user21820, Paul Frost, amWhy, T. Bongers, José Carlos Santos yesterday
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." – user21820, Paul Frost, amWhy, T. Bongers, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
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show 3 more comments
up vote
9
down vote
favorite
11
Find integer numbers $m$, $n$ such that $m^{5}n+7$ and $n^{5}m+7$ are cubes of integer.
I guess the only pair is $(1;1)$, but I don't know what way we should follow to solve this problem. Help me please.
number-theory elementary-number-theory diophantine-equations problem-solving
edited Nov 14 at 15:03
Batominovski
31.5k23187
asked Nov 14 at 14:22
Trong Tuan
585
put on hold as off-topic by user21820, Paul Frost, amWhy, T. Bongers, José Carlos Santos yesterday
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." – user21820, Paul Frost, amWhy, T. Bongers, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
3
I think you should specify the source of this problem. Where did you find it?
– астон вілла олоф мэллбэрг
Nov 14 at 14:40
Looks like an Olympiad problem doesn't it ...
– Richard Martin
Nov 14 at 17:03
1
artofproblemsolving.com/community/c5h532232p3041819 but it doesn't help since here we have a solution
– greedoid
Nov 14 at 17:04
Isn'it already known that $M^2+7$ is cube iff $M=1$? Then, it would suffice to show that under these two conditions, we must have $m=n$.
– René Gy
Nov 14 at 18:13
1
@Servaes Your right: my idea goes nowhere. All the more that my assumption $M^2+7$ is cube iff $M=1$ is just wrong, since $181^2+7=32^3$..
– René Gy
Nov 14 at 18:47
|
show 3 more comments
up vote
9
down vote
favorite
11
up vote
9
down vote
favorite
11
11
Find integer numbers $m$, $n$ such that $m^{5}n+7$ and $n^{5}m+7$ are cubes of integer.
I guess the only pair is $(1;1)$, but I don't know what way we should follow to solve this problem. Help me please.
number-theory elementary-number-theory diophantine-equations problem-solving
edited Nov 14 at 15:03
Batominovski
31.5k23187
asked Nov 14 at 14:22
Trong Tuan
585
Find integer numbers $m$, $n$ such that $m^{5}n+7$ and $n^{5}m+7$ are cubes of integer.
I guess the only pair is $(1;1)$, but I don't know what way we should follow to solve this problem. Help me please.
number-theory elementary-number-theory diophantine-equations problem-solving
number-theory elementary-number-theory diophantine-equations problem-solving
edited Nov 14 at 15:03
Batominovski
31.5k23187
asked Nov 14 at 14:22
Trong Tuan
585
edited Nov 14 at 15:03
Batominovski
31.5k23187
asked Nov 14 at 14:22
Trong Tuan
585
edited Nov 14 at 15:03
Batominovski
31.5k23187
edited Nov 14 at 15:03
Batominovski
31.5k23187
edited Nov 14 at 15:03
Batominovski
31.5k23187
31.5k23187
asked Nov 14 at 14:22
Trong Tuan
585
asked Nov 14 at 14:22
Trong Tuan
585
asked Nov 14 at 14:22
Trong Tuan
585
585
put on hold as off-topic by user21820, Paul Frost, amWhy, T. Bongers, José Carlos Santos yesterday
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." – user21820, Paul Frost, amWhy, T. Bongers, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as off-topic by user21820, Paul Frost, amWhy, T. Bongers, José Carlos Santos yesterday
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." – user21820, Paul Frost, amWhy, T. Bongers, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
3
I think you should specify the source of this problem. Where did you find it?
– астон вілла олоф мэллбэрг
Nov 14 at 14:40
Looks like an Olympiad problem doesn't it ...
– Richard Martin
Nov 14 at 17:03
1
artofproblemsolving.com/community/c5h532232p3041819 but it doesn't help since here we have a solution
– greedoid
Nov 14 at 17:04
Isn'it already known that $M^2+7$ is cube iff $M=1$? Then, it would suffice to show that under these two conditions, we must have $m=n$.
– René Gy
Nov 14 at 18:13
1
@Servaes Your right: my idea goes nowhere. All the more that my assumption $M^2+7$ is cube iff $M=1$ is just wrong, since $181^2+7=32^3$..
– René Gy
Nov 14 at 18:47
|
show 3 more comments
3
I think you should specify the source of this problem. Where did you find it?
– астон вілла олоф мэллбэрг
Nov 14 at 14:40
Looks like an Olympiad problem doesn't it ...
– Richard Martin
Nov 14 at 17:03
1
artofproblemsolving.com/community/c5h532232p3041819 but it doesn't help since here we have a solution
– greedoid
Nov 14 at 17:04
Isn'it already known that $M^2+7$ is cube iff $M=1$? Then, it would suffice to show that under these two conditions, we must have $m=n$.
– René Gy
Nov 14 at 18:13
1
@Servaes Your right: my idea goes nowhere. All the more that my assumption $M^2+7$ is cube iff $M=1$ is just wrong, since $181^2+7=32^3$..
– René Gy
Nov 14 at 18:47
3
3
I think you should specify the source of this problem. Where did you find it?
– астон вілла олоф мэллбэрг
Nov 14 at 14:40
I think you should specify the source of this problem. Where did you find it?
– астон вілла олоф мэллбэрг
Nov 14 at 14:40
Looks like an Olympiad problem doesn't it ...
– Richard Martin
Nov 14 at 17:03
Looks like an Olympiad problem doesn't it ...
– Richard Martin
Nov 14 at 17:03
1
1
artofproblemsolving.com/community/c5h532232p3041819 but it doesn't help since here we have a solution
– greedoid
Nov 14 at 17:04
artofproblemsolving.com/community/c5h532232p3041819 but it doesn't help since here we have a solution
– greedoid
Nov 14 at 17:04
Isn'it already known that $M^2+7$ is cube iff $M=1$? Then, it would suffice to show that under these two conditions, we must have $m=n$.
– René Gy
Nov 14 at 18:13
Isn'it already known that $M^2+7$ is cube iff $M=1$? Then, it would suffice to show that under these two conditions, we must have $m=n$.
– René Gy
Nov 14 at 18:13
1
1
@Servaes Your right: my idea goes nowhere. All the more that my assumption $M^2+7$ is cube iff $M=1$ is just wrong, since $181^2+7=32^3$..
– René Gy
Nov 14 at 18:47
@Servaes Your right: my idea goes nowhere. All the more that my assumption $M^2+7$ is cube iff $M=1$ is just wrong, since $181^2+7=32^3$..
– René Gy
Nov 14 at 18:47
|
show 3 more comments
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3
I think you should specify the source of this problem. Where did you find it?
– астон вілла олоф мэллбэрг
Nov 14 at 14:40
Looks like an Olympiad problem doesn't it ...
– Richard Martin
Nov 14 at 17:03
1
artofproblemsolving.com/community/c5h532232p3041819 but it doesn't help since here we have a solution
– greedoid
Nov 14 at 17:04
Isn'it already known that $M^2+7$ is cube iff $M=1$? Then, it would suffice to show that under these two conditions, we must have $m=n$.
– René Gy
Nov 14 at 18:13
1
@Servaes Your right: my idea goes nowhere. All the more that my assumption $M^2+7$ is cube iff $M=1$ is just wrong, since $181^2+7=32^3$..
– René Gy
Nov 14 at 18:47