Mixing
Resolving a rational system of equations with too many unkowns
$begingroup$
A little bit of context.
While working a larger proof (the proof is quite related to this question I asked), I stumbled upon the following problem.
The question.
Can we find $x_1,ldots,x_{24}inmathbb Q$ such that for all $a,b,c,dinmathbb Q$:
$$ (x_1a+x_2b+x_3c+x_4d)(x_5a+x_6b+x_7c+x_8d)-(x_{9}a+x_{10}b+x_{11}c+x_{12}d)(x_{13}a+x_{14}b+x_{15}c+x_{16}d)-(x_{17}a+x_{18}b+x_{19}c+x_{20}d)(x_{21}a+x_{22}b+x_{23}c+x_{24}d)$$
$$=7a^2-b^2-c^2-d^2quad ?$$
What I tried.
I have tried to do a systematic resolution of the underlying system, but this is too massive to compute.
I have tried arbitrarily fixing some of the $x_i$ in order to simplify the system, but I always end up in one of the following cases:
no solution,
irrational solutions,
complex solutions.
None of this cases is acceptable.
Any leads would be greatly appreciated.
algebra-precalculus systems-of-equations diophantine-equations
asked Jan 21 at 15:03
E. JosephE. Joseph
11.7k82856
$endgroup$
add a comment |
$begingroup$
A little bit of context.
While working a larger proof (the proof is quite related to this question I asked), I stumbled upon the following problem.
The question.
Can we find $x_1,ldots,x_{24}inmathbb Q$ such that for all $a,b,c,dinmathbb Q$:
$$ (x_1a+x_2b+x_3c+x_4d)(x_5a+x_6b+x_7c+x_8d)-(x_{9}a+x_{10}b+x_{11}c+x_{12}d)(x_{13}a+x_{14}b+x_{15}c+x_{16}d)-(x_{17}a+x_{18}b+x_{19}c+x_{20}d)(x_{21}a+x_{22}b+x_{23}c+x_{24}d)$$
$$=7a^2-b^2-c^2-d^2quad ?$$
What I tried.
I have tried to do a systematic resolution of the underlying system, but this is too massive to compute.
I have tried arbitrarily fixing some of the $x_i$ in order to simplify the system, but I always end up in one of the following cases:
no solution,
irrational solutions,
complex solutions.
None of this cases is acceptable.
Any leads would be greatly appreciated.
algebra-precalculus systems-of-equations diophantine-equations
asked Jan 21 at 15:03
E. JosephE. Joseph
11.7k82856
$endgroup$
$begingroup$
You can just remove all the terms $x_{17}$ through $x_{24}$. They are redundant with the preceding eight. Replace $x_9$ with $x_9+x_{17}$ and so on.
$endgroup$
– Ross Millikan
Jan 21 at 15:18
add a comment |
$begingroup$
A little bit of context.
While working a larger proof (the proof is quite related to this question I asked), I stumbled upon the following problem.
The question.
Can we find $x_1,ldots,x_{24}inmathbb Q$ such that for all $a,b,c,dinmathbb Q$:
$$ (x_1a+x_2b+x_3c+x_4d)(x_5a+x_6b+x_7c+x_8d)-(x_{9}a+x_{10}b+x_{11}c+x_{12}d)(x_{13}a+x_{14}b+x_{15}c+x_{16}d)-(x_{17}a+x_{18}b+x_{19}c+x_{20}d)(x_{21}a+x_{22}b+x_{23}c+x_{24}d)$$
$$=7a^2-b^2-c^2-d^2quad ?$$
What I tried.
I have tried to do a systematic resolution of the underlying system, but this is too massive to compute.
I have tried arbitrarily fixing some of the $x_i$ in order to simplify the system, but I always end up in one of the following cases:
no solution,
irrational solutions,
complex solutions.
None of this cases is acceptable.
Any leads would be greatly appreciated.
algebra-precalculus systems-of-equations diophantine-equations
asked Jan 21 at 15:03
E. JosephE. Joseph
11.7k82856
$endgroup$
A little bit of context.
While working a larger proof (the proof is quite related to this question I asked), I stumbled upon the following problem.
The question.
Can we find $x_1,ldots,x_{24}inmathbb Q$ such that for all $a,b,c,dinmathbb Q$:
$$ (x_1a+x_2b+x_3c+x_4d)(x_5a+x_6b+x_7c+x_8d)-(x_{9}a+x_{10}b+x_{11}c+x_{12}d)(x_{13}a+x_{14}b+x_{15}c+x_{16}d)-(x_{17}a+x_{18}b+x_{19}c+x_{20}d)(x_{21}a+x_{22}b+x_{23}c+x_{24}d)$$
$$=7a^2-b^2-c^2-d^2quad ?$$
What I tried.
I have tried to do a systematic resolution of the underlying system, but this is too massive to compute.
I have tried arbitrarily fixing some of the $x_i$ in order to simplify the system, but I always end up in one of the following cases:
no solution,
irrational solutions,
complex solutions.
None of this cases is acceptable.
Any leads would be greatly appreciated.
algebra-precalculus systems-of-equations diophantine-equations
algebra-precalculus systems-of-equations diophantine-equations
asked Jan 21 at 15:03
E. JosephE. Joseph
11.7k82856
asked Jan 21 at 15:03
E. JosephE. Joseph
11.7k82856
asked Jan 21 at 15:03
E. JosephE. Joseph
11.7k82856
asked Jan 21 at 15:03
E. JosephE. Joseph
11.7k82856
asked Jan 21 at 15:03
E. JosephE. Joseph
11.7k82856
11.7k82856
$begingroup$
You can just remove all the terms $x_{17}$ through $x_{24}$. They are redundant with the preceding eight. Replace $x_9$ with $x_9+x_{17}$ and so on.
$endgroup$
– Ross Millikan
Jan 21 at 15:18
add a comment |
$begingroup$
You can just remove all the terms $x_{17}$ through $x_{24}$. They are redundant with the preceding eight. Replace $x_9$ with $x_9+x_{17}$ and so on.
$endgroup$
– Ross Millikan
Jan 21 at 15:18
$begingroup$
You can just remove all the terms $x_{17}$ through $x_{24}$. They are redundant with the preceding eight. Replace $x_9$ with $x_9+x_{17}$ and so on.
$endgroup$
– Ross Millikan
Jan 21 at 15:18
$begingroup$
You can just remove all the terms $x_{17}$ through $x_{24}$. They are redundant with the preceding eight. Replace $x_9$ with $x_9+x_{17}$ and so on.
$endgroup$
– Ross Millikan
Jan 21 at 15:18
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
From Legendre $3$-square theorem the right hand side can never vanish unless all terms are zero.
However, no mather what the $x_i$ are, it is possible with three linear conditions on $(a,b,c,d)$ to force the left hand side to vanish, meaning the left hand side has infinitely many rational zeroes.
answered Jan 21 at 15:17
MindlackMindlack
4,885211
$endgroup$
$begingroup$
Thanks for the answer. Can we find explicitly these linear conditions on $(a,b,c,d,)$?
$endgroup$
– E. Joseph
Jan 22 at 10:46
$begingroup$
Vanish one factor per term in your left hand side.
$endgroup$
– Mindlack
Jan 22 at 10:47
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
From Legendre $3$-square theorem the right hand side can never vanish unless all terms are zero.
However, no mather what the $x_i$ are, it is possible with three linear conditions on $(a,b,c,d)$ to force the left hand side to vanish, meaning the left hand side has infinitely many rational zeroes.
answered Jan 21 at 15:17
MindlackMindlack
4,885211
$endgroup$
$begingroup$
Thanks for the answer. Can we find explicitly these linear conditions on $(a,b,c,d,)$?
$endgroup$
– E. Joseph
Jan 22 at 10:46
$begingroup$
Vanish one factor per term in your left hand side.
$endgroup$
– Mindlack
Jan 22 at 10:47
add a comment |
$begingroup$
From Legendre $3$-square theorem the right hand side can never vanish unless all terms are zero.
However, no mather what the $x_i$ are, it is possible with three linear conditions on $(a,b,c,d)$ to force the left hand side to vanish, meaning the left hand side has infinitely many rational zeroes.
answered Jan 21 at 15:17
MindlackMindlack
4,885211
$endgroup$
$begingroup$
Thanks for the answer. Can we find explicitly these linear conditions on $(a,b,c,d,)$?
$endgroup$
– E. Joseph
Jan 22 at 10:46
$begingroup$
Vanish one factor per term in your left hand side.
$endgroup$
– Mindlack
Jan 22 at 10:47
add a comment |
$begingroup$
From Legendre $3$-square theorem the right hand side can never vanish unless all terms are zero.
However, no mather what the $x_i$ are, it is possible with three linear conditions on $(a,b,c,d)$ to force the left hand side to vanish, meaning the left hand side has infinitely many rational zeroes.
answered Jan 21 at 15:17
MindlackMindlack
4,885211
$endgroup$
From Legendre $3$-square theorem the right hand side can never vanish unless all terms are zero.
However, no mather what the $x_i$ are, it is possible with three linear conditions on $(a,b,c,d)$ to force the left hand side to vanish, meaning the left hand side has infinitely many rational zeroes.
answered Jan 21 at 15:17
MindlackMindlack
4,885211
answered Jan 21 at 15:17
MindlackMindlack
4,885211
answered Jan 21 at 15:17
MindlackMindlack
4,885211
answered Jan 21 at 15:17
MindlackMindlack
4,885211
4,885211
$begingroup$
Thanks for the answer. Can we find explicitly these linear conditions on $(a,b,c,d,)$?
$endgroup$
– E. Joseph
Jan 22 at 10:46
$begingroup$
Vanish one factor per term in your left hand side.
$endgroup$
– Mindlack
Jan 22 at 10:47
add a comment |
$begingroup$
Thanks for the answer. Can we find explicitly these linear conditions on $(a,b,c,d,)$?
$endgroup$
– E. Joseph
Jan 22 at 10:46
$begingroup$
Vanish one factor per term in your left hand side.
$endgroup$
– Mindlack
Jan 22 at 10:47
$begingroup$
Thanks for the answer. Can we find explicitly these linear conditions on $(a,b,c,d,)$?
$endgroup$
– E. Joseph
Jan 22 at 10:46
$begingroup$
Thanks for the answer. Can we find explicitly these linear conditions on $(a,b,c,d,)$?
$endgroup$
– E. Joseph
Jan 22 at 10:46
$begingroup$
Vanish one factor per term in your left hand side.
$endgroup$
– Mindlack
Jan 22 at 10:47
$begingroup$
Vanish one factor per term in your left hand side.
$endgroup$
– Mindlack
Jan 22 at 10:47
add a comment |
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$begingroup$
You can just remove all the terms $x_{17}$ through $x_{24}$. They are redundant with the preceding eight. Replace $x_9$ with $x_9+x_{17}$ and so on.
$endgroup$
– Ross Millikan
Jan 21 at 15:18