Mixing
How to solve $a^3+2b^3+6c^3=380$ for $a, b, c in mathbb{Z}$ [closed]
2
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According to computer calculations, the integer solution is $(x, y, z)=(6, 1, 3)$ only.
number-theory
asked Jan 24 at 14:54
user458792user458792
113
$endgroup$
closed as off-topic by Namaste, Crostul, José Carlos Santos, metamorphy, Cesareo Jan 25 at 10:20
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." – Namaste, Crostul, José Carlos Santos, metamorphy, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
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2
$begingroup$
According to computer calculations, the integer solution is $(x, y, z)=(6, 1, 3)$ only.
number-theory
asked Jan 24 at 14:54
user458792user458792
113
$endgroup$
closed as off-topic by Namaste, Crostul, José Carlos Santos, metamorphy, Cesareo Jan 25 at 10:20
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." – Namaste, Crostul, José Carlos Santos, metamorphy, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
I hardly believe that this equation can be solved in an elementary way. There may be some solution with $a,b,c$ very large.
$endgroup$
– Crostul
Jan 24 at 14:58
1
$begingroup$
@Crostul it's an elliptic curve, so there will be some algorithm to find all solutions. I don't want to do it, though.
$endgroup$
– hunter
Jan 24 at 15:05
1
$begingroup$
By the way, other solutions are $(a,b,c)=(18,-17,9)$ and $(a,b,c)=(60,1,-33)$, found with brute force.
$endgroup$
– Crostul
Jan 24 at 15:06
1
$begingroup$
I think (6,1,3) is the only solution with positive integers. If you only want positive solutions, they can be found with elementary means.
$endgroup$
– quarague
Jan 24 at 15:18
1
$begingroup$
If only positive solutions are allowed, the problem is trivial and can quickly be done by brute force. The hard part is to find all solutions over the integers. Can this really be solved with methods of elliptic curves although the exponent $3$ appears three times ?
$endgroup$
– Peter
Jan 24 at 18:39
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show 6 more comments
2
2
2
2
$begingroup$
According to computer calculations, the integer solution is $(x, y, z)=(6, 1, 3)$ only.
number-theory
asked Jan 24 at 14:54
user458792user458792
113
$endgroup$
According to computer calculations, the integer solution is $(x, y, z)=(6, 1, 3)$ only.
number-theory
number-theory
asked Jan 24 at 14:54
user458792user458792
113
asked Jan 24 at 14:54
user458792user458792
113
edited Jan 24 at 16:25
user458792
asked Jan 24 at 14:54
user458792user458792
113
asked Jan 24 at 14:54
user458792user458792
113
asked Jan 24 at 14:54
user458792user458792
113
113
closed as off-topic by Namaste, Crostul, José Carlos Santos, metamorphy, Cesareo Jan 25 at 10:20
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." – Namaste, Crostul, José Carlos Santos, metamorphy, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Namaste, Crostul, José Carlos Santos, metamorphy, Cesareo Jan 25 at 10:20
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." – Namaste, Crostul, José Carlos Santos, metamorphy, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
I hardly believe that this equation can be solved in an elementary way. There may be some solution with $a,b,c$ very large.
$endgroup$
– Crostul
Jan 24 at 14:58
1
$begingroup$
@Crostul it's an elliptic curve, so there will be some algorithm to find all solutions. I don't want to do it, though.
$endgroup$
– hunter
Jan 24 at 15:05
1
$begingroup$
By the way, other solutions are $(a,b,c)=(18,-17,9)$ and $(a,b,c)=(60,1,-33)$, found with brute force.
$endgroup$
– Crostul
Jan 24 at 15:06
1
$begingroup$
I think (6,1,3) is the only solution with positive integers. If you only want positive solutions, they can be found with elementary means.
$endgroup$
– quarague
Jan 24 at 15:18
1
$begingroup$
If only positive solutions are allowed, the problem is trivial and can quickly be done by brute force. The hard part is to find all solutions over the integers. Can this really be solved with methods of elliptic curves although the exponent $3$ appears three times ?
$endgroup$
– Peter
Jan 24 at 18:39
|
show 6 more comments
1
$begingroup$
I hardly believe that this equation can be solved in an elementary way. There may be some solution with $a,b,c$ very large.
$endgroup$
– Crostul
Jan 24 at 14:58
1
$begingroup$
@Crostul it's an elliptic curve, so there will be some algorithm to find all solutions. I don't want to do it, though.
$endgroup$
– hunter
Jan 24 at 15:05
1
$begingroup$
By the way, other solutions are $(a,b,c)=(18,-17,9)$ and $(a,b,c)=(60,1,-33)$, found with brute force.
$endgroup$
– Crostul
Jan 24 at 15:06
1
$begingroup$
I think (6,1,3) is the only solution with positive integers. If you only want positive solutions, they can be found with elementary means.
$endgroup$
– quarague
Jan 24 at 15:18
1
$begingroup$
If only positive solutions are allowed, the problem is trivial and can quickly be done by brute force. The hard part is to find all solutions over the integers. Can this really be solved with methods of elliptic curves although the exponent $3$ appears three times ?
$endgroup$
– Peter
Jan 24 at 18:39
1
1
$begingroup$
I hardly believe that this equation can be solved in an elementary way. There may be some solution with $a,b,c$ very large.
$endgroup$
– Crostul
Jan 24 at 14:58
$begingroup$
I hardly believe that this equation can be solved in an elementary way. There may be some solution with $a,b,c$ very large.
$endgroup$
– Crostul
Jan 24 at 14:58
1
1
$begingroup$
@Crostul it's an elliptic curve, so there will be some algorithm to find all solutions. I don't want to do it, though.
$endgroup$
– hunter
Jan 24 at 15:05
$begingroup$
@Crostul it's an elliptic curve, so there will be some algorithm to find all solutions. I don't want to do it, though.
$endgroup$
– hunter
Jan 24 at 15:05
1
1
$begingroup$
By the way, other solutions are $(a,b,c)=(18,-17,9)$ and $(a,b,c)=(60,1,-33)$, found with brute force.
$endgroup$
– Crostul
Jan 24 at 15:06
$begingroup$
By the way, other solutions are $(a,b,c)=(18,-17,9)$ and $(a,b,c)=(60,1,-33)$, found with brute force.
$endgroup$
– Crostul
Jan 24 at 15:06
1
1
$begingroup$
I think (6,1,3) is the only solution with positive integers. If you only want positive solutions, they can be found with elementary means.
$endgroup$
– quarague
Jan 24 at 15:18
$begingroup$
I think (6,1,3) is the only solution with positive integers. If you only want positive solutions, they can be found with elementary means.
$endgroup$
– quarague
Jan 24 at 15:18
1
1
$begingroup$
If only positive solutions are allowed, the problem is trivial and can quickly be done by brute force. The hard part is to find all solutions over the integers. Can this really be solved with methods of elliptic curves although the exponent $3$ appears three times ?
$endgroup$
– Peter
Jan 24 at 18:39
$begingroup$
If only positive solutions are allowed, the problem is trivial and can quickly be done by brute force. The hard part is to find all solutions over the integers. Can this really be solved with methods of elliptic curves although the exponent $3$ appears three times ?
$endgroup$
– Peter
Jan 24 at 18:39
|
show 6 more comments
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1
$begingroup$
I hardly believe that this equation can be solved in an elementary way. There may be some solution with $a,b,c$ very large.
$endgroup$
– Crostul
Jan 24 at 14:58
1
$begingroup$
@Crostul it's an elliptic curve, so there will be some algorithm to find all solutions. I don't want to do it, though.
$endgroup$
– hunter
Jan 24 at 15:05
1
$begingroup$
By the way, other solutions are $(a,b,c)=(18,-17,9)$ and $(a,b,c)=(60,1,-33)$, found with brute force.
$endgroup$
– Crostul
Jan 24 at 15:06
1
$begingroup$
I think (6,1,3) is the only solution with positive integers. If you only want positive solutions, they can be found with elementary means.
$endgroup$
– quarague
Jan 24 at 15:18
1
$begingroup$
If only positive solutions are allowed, the problem is trivial and can quickly be done by brute force. The hard part is to find all solutions over the integers. Can this really be solved with methods of elliptic curves although the exponent $3$ appears three times ?
$endgroup$
– Peter
Jan 24 at 18:39