Rational solution to a system of equations
$begingroup$
Some context.
By rational subspace, I mean a subspace of $mathbb R^5$ which admits a rational basis. In other words, a basis formed with vectors of $mathbb Q^5$.
For instance, the vector $v=(0,pi,3pi,pi-pi^2)$ is in a rational subspace of dimension $2$ since:
$$v=pibegin{pmatrix} 0 \ 1 \ 3 \ 1end{pmatrix}+pi^2begin{pmatrix} 0 \ 0 \ 0 \ -1end{pmatrix}.$$
The question.
Let $A=mathrm{Span}(Y_1,Y_2,Y_3)$ where
$$Y_1=begin{pmatrix} 1 \ 0 \ 0 \ sqrt 6 \ sqrt {15} end{pmatrix},quad Y_2=begin{pmatrix} 0 \ sqrt{6} \ 0 \ -sqrt {15} \ sqrt {10} end{pmatrix}quadtext { and }quad Y_3=begin{pmatrix} 0 \ 0 \ sqrt{15} \ sqrt {10} \ sqrt {6} end{pmatrix}.$$
Does there exist a rational subspace $B$ of $mathbb R^5$, of dimension $2$, such that $Acap Bne{0}$?
I believe the answer to be negative.
What I tried.
For $Yin A$, let's denote by $Y_1,ldots,Y_5inmathbb R$ its coordinates. We can prove the following lemma.
Lemma. The answer to the question is negative, if, and only if,
$$forall Y in A,quad dim_{mathbb Q}(mathrm{Span}_{mathbb Q}(Y_1,ldots,Y_5)ge 3.$$
I tried to investigate what such a rational subspace $B$ would look like. Let's take $B$, a rational subspace of $mathbb R^5$ of dimension $2$, such that $Acap Bne{0}$.
Let $Y:=alpha Y_1+beta Y_2+gamma Y_3$ be a vector in $Asetminus{0}$.
We have
$$alpha Y_1+beta Y_2+gamma Y_3=begin{pmatrix} alpha \ betasqrt 6 \ gamma sqrt {15} \ alphasqrt 6-betasqrt {15} +gammasqrt{10} \ alphasqrt{15}+betasqrt{10}+gammasqrt{6}end{pmatrix}.$$
If we assume (to try to get somewhere) that
$$dim_{mathbb Q}(mathrm{Span}_{mathbb Q}(alpha,sqrt{6}beta)=2,$$
then the last three coordinates of $Y$ must be a rational linear combination of the first two, i.e. there exists $x_1,ldots,x_6inmathbb Q$ such that
$$begin{cases}gamma sqrt {15}=x_1alpha+x_2betasqrt 6 \ alphasqrt 6-betasqrt {15} +gammasqrt{10}= x_3alpha+x_4betasqrt 6 \ alphasqrt{15}+betasqrt{10}+gammasqrt{6}=x_5alpha+x_6betasqrt 6end{cases}$$
which can be rewrite
$$MX=0quadtext{ with } quad M=begin{pmatrix} -x_1 & -x_2sqrt 6 & sqrt{15} \ sqrt{6}-x_3 & -sqrt{15}-x_4sqrt 6 & sqrt{10} \ sqrt{15}-x_5 & sqrt{10}- x_6sqrt 6 & sqrt 6 end{pmatrix}quad text{ and }quad X=begin{pmatrix} alpha \ beta\ gammaend{pmatrix}.$$
So if $det(M)ne 0$, we have won, since $(alpha,beta,gamma)=(0,0,0)$ is the only solution, thus $Y=0$ which is absurd.
Let's compute $det(M)$ then:
$$det(M)=Asqrt{15}+Bsqrt{10}+Csqrt 6+D,$$
with $A,B,C,Dinmathbb Q$ depending on the $x_i$. Since
$$dim_{mathbb Q}(mathrm{Span}_{mathbb Q}(sqrt 6,sqrt{10},sqrt{15}))=3,$$
and $A,B,C,Dinmathbb Q$, if we assume $det(M)=0$, we must have
$$A=B=C=D=0.$$
The computations give
$$(mathscr S)quad begin{cases} 2x_2x_5-2x_1x_6-6x_6+15=0 \ -3x_4x_5+3x_3x_6+3x_1=0 \ 6x_2-5x_3+15x_4=0 \ 6x_2x_3+6x_1x_4+10x_1-30x_2-15x_5+30=0.end{cases}$$
The question can now be reformulated as follow:
Does the system $(mathscr S)$ has rational solutions?
If we try to solve the system, we can end up with this expression:
$frac{20 , x_{1} x_{3}^{2} x_{5} + 12 , x_{1}^{3} - 30 , x_{1}^{2} x_{3} + 20 , x_{1}^{2} x_{5} + 30 , x_{3}^{2} x_{5} - 30 , x_{1} x_{5}^{2} - 75 , x_{3} x_{5}^{2} + 186 , x_{1}^{2} - 225 , x_{3}^{2} + 120 , x_{1} x_{5} - 90 , x_{5}^{2} + 450 , x_{1} + 1125 , x_{3} + 180 , x_{5}}{2 , x_{1} x_{5} + 5 , x_{3} x_{5} + 6 , x_{5}}=0.$
The question is then to understand if this surface in $mathbb R^3$ has rational points. If you are curious about it, this is what the numerator looks like:
Final remarks.
This question really interests me, but I feel quite stuck about it. May be my whole approach isn't going to help. Any hints, references or piece of solutions would be much appreciated.
linear-algebra number-theory vector-spaces diophantine-equations rational-numbers
asked Dec 1 '18 at 14:31
E. JosephE. Joseph
11.7k82856
$endgroup$
add a comment |
$begingroup$
Some context.
By rational subspace, I mean a subspace of $mathbb R^5$ which admits a rational basis. In other words, a basis formed with vectors of $mathbb Q^5$.
For instance, the vector $v=(0,pi,3pi,pi-pi^2)$ is in a rational subspace of dimension $2$ since:
$$v=pibegin{pmatrix} 0 \ 1 \ 3 \ 1end{pmatrix}+pi^2begin{pmatrix} 0 \ 0 \ 0 \ -1end{pmatrix}.$$
The question.
Let $A=mathrm{Span}(Y_1,Y_2,Y_3)$ where
$$Y_1=begin{pmatrix} 1 \ 0 \ 0 \ sqrt 6 \ sqrt {15} end{pmatrix},quad Y_2=begin{pmatrix} 0 \ sqrt{6} \ 0 \ -sqrt {15} \ sqrt {10} end{pmatrix}quadtext { and }quad Y_3=begin{pmatrix} 0 \ 0 \ sqrt{15} \ sqrt {10} \ sqrt {6} end{pmatrix}.$$
Does there exist a rational subspace $B$ of $mathbb R^5$, of dimension $2$, such that $Acap Bne{0}$?
I believe the answer to be negative.
What I tried.
For $Yin A$, let's denote by $Y_1,ldots,Y_5inmathbb R$ its coordinates. We can prove the following lemma.
Lemma. The answer to the question is negative, if, and only if,
$$forall Y in A,quad dim_{mathbb Q}(mathrm{Span}_{mathbb Q}(Y_1,ldots,Y_5)ge 3.$$
I tried to investigate what such a rational subspace $B$ would look like. Let's take $B$, a rational subspace of $mathbb R^5$ of dimension $2$, such that $Acap Bne{0}$.
Let $Y:=alpha Y_1+beta Y_2+gamma Y_3$ be a vector in $Asetminus{0}$.
We have
$$alpha Y_1+beta Y_2+gamma Y_3=begin{pmatrix} alpha \ betasqrt 6 \ gamma sqrt {15} \ alphasqrt 6-betasqrt {15} +gammasqrt{10} \ alphasqrt{15}+betasqrt{10}+gammasqrt{6}end{pmatrix}.$$
If we assume (to try to get somewhere) that
$$dim_{mathbb Q}(mathrm{Span}_{mathbb Q}(alpha,sqrt{6}beta)=2,$$
then the last three coordinates of $Y$ must be a rational linear combination of the first two, i.e. there exists $x_1,ldots,x_6inmathbb Q$ such that
$$begin{cases}gamma sqrt {15}=x_1alpha+x_2betasqrt 6 \ alphasqrt 6-betasqrt {15} +gammasqrt{10}= x_3alpha+x_4betasqrt 6 \ alphasqrt{15}+betasqrt{10}+gammasqrt{6}=x_5alpha+x_6betasqrt 6end{cases}$$
which can be rewrite
$$MX=0quadtext{ with } quad M=begin{pmatrix} -x_1 & -x_2sqrt 6 & sqrt{15} \ sqrt{6}-x_3 & -sqrt{15}-x_4sqrt 6 & sqrt{10} \ sqrt{15}-x_5 & sqrt{10}- x_6sqrt 6 & sqrt 6 end{pmatrix}quad text{ and }quad X=begin{pmatrix} alpha \ beta\ gammaend{pmatrix}.$$
So if $det(M)ne 0$, we have won, since $(alpha,beta,gamma)=(0,0,0)$ is the only solution, thus $Y=0$ which is absurd.
Let's compute $det(M)$ then:
$$det(M)=Asqrt{15}+Bsqrt{10}+Csqrt 6+D,$$
with $A,B,C,Dinmathbb Q$ depending on the $x_i$. Since
$$dim_{mathbb Q}(mathrm{Span}_{mathbb Q}(sqrt 6,sqrt{10},sqrt{15}))=3,$$
and $A,B,C,Dinmathbb Q$, if we assume $det(M)=0$, we must have
$$A=B=C=D=0.$$
The computations give
$$(mathscr S)quad begin{cases} 2x_2x_5-2x_1x_6-6x_6+15=0 \ -3x_4x_5+3x_3x_6+3x_1=0 \ 6x_2-5x_3+15x_4=0 \ 6x_2x_3+6x_1x_4+10x_1-30x_2-15x_5+30=0.end{cases}$$
The question can now be reformulated as follow:
Does the system $(mathscr S)$ has rational solutions?
If we try to solve the system, we can end up with this expression:
$frac{20 , x_{1} x_{3}^{2} x_{5} + 12 , x_{1}^{3} - 30 , x_{1}^{2} x_{3} + 20 , x_{1}^{2} x_{5} + 30 , x_{3}^{2} x_{5} - 30 , x_{1} x_{5}^{2} - 75 , x_{3} x_{5}^{2} + 186 , x_{1}^{2} - 225 , x_{3}^{2} + 120 , x_{1} x_{5} - 90 , x_{5}^{2} + 450 , x_{1} + 1125 , x_{3} + 180 , x_{5}}{2 , x_{1} x_{5} + 5 , x_{3} x_{5} + 6 , x_{5}}=0.$
The question is then to understand if this surface in $mathbb R^3$ has rational points. If you are curious about it, this is what the numerator looks like:
Final remarks.
This question really interests me, but I feel quite stuck about it. May be my whole approach isn't going to help. Any hints, references or piece of solutions would be much appreciated.
linear-algebra number-theory vector-spaces diophantine-equations rational-numbers
asked Dec 1 '18 at 14:31
E. JosephE. Joseph
11.7k82856
$endgroup$
1
$begingroup$
@AlexRavsky The lemma was indeed maybe confusing, so I made it a little more clear I hope.
$endgroup$
– E. Joseph
Jan 13 at 9:25
add a comment |
$begingroup$
Some context.
By rational subspace, I mean a subspace of $mathbb R^5$ which admits a rational basis. In other words, a basis formed with vectors of $mathbb Q^5$.
For instance, the vector $v=(0,pi,3pi,pi-pi^2)$ is in a rational subspace of dimension $2$ since:
$$v=pibegin{pmatrix} 0 \ 1 \ 3 \ 1end{pmatrix}+pi^2begin{pmatrix} 0 \ 0 \ 0 \ -1end{pmatrix}.$$
The question.
Let $A=mathrm{Span}(Y_1,Y_2,Y_3)$ where
$$Y_1=begin{pmatrix} 1 \ 0 \ 0 \ sqrt 6 \ sqrt {15} end{pmatrix},quad Y_2=begin{pmatrix} 0 \ sqrt{6} \ 0 \ -sqrt {15} \ sqrt {10} end{pmatrix}quadtext { and }quad Y_3=begin{pmatrix} 0 \ 0 \ sqrt{15} \ sqrt {10} \ sqrt {6} end{pmatrix}.$$
Does there exist a rational subspace $B$ of $mathbb R^5$, of dimension $2$, such that $Acap Bne{0}$?
I believe the answer to be negative.
What I tried.
For $Yin A$, let's denote by $Y_1,ldots,Y_5inmathbb R$ its coordinates. We can prove the following lemma.
Lemma. The answer to the question is negative, if, and only if,
$$forall Y in A,quad dim_{mathbb Q}(mathrm{Span}_{mathbb Q}(Y_1,ldots,Y_5)ge 3.$$
I tried to investigate what such a rational subspace $B$ would look like. Let's take $B$, a rational subspace of $mathbb R^5$ of dimension $2$, such that $Acap Bne{0}$.
Let $Y:=alpha Y_1+beta Y_2+gamma Y_3$ be a vector in $Asetminus{0}$.
We have
$$alpha Y_1+beta Y_2+gamma Y_3=begin{pmatrix} alpha \ betasqrt 6 \ gamma sqrt {15} \ alphasqrt 6-betasqrt {15} +gammasqrt{10} \ alphasqrt{15}+betasqrt{10}+gammasqrt{6}end{pmatrix}.$$
If we assume (to try to get somewhere) that
$$dim_{mathbb Q}(mathrm{Span}_{mathbb Q}(alpha,sqrt{6}beta)=2,$$
then the last three coordinates of $Y$ must be a rational linear combination of the first two, i.e. there exists $x_1,ldots,x_6inmathbb Q$ such that
$$begin{cases}gamma sqrt {15}=x_1alpha+x_2betasqrt 6 \ alphasqrt 6-betasqrt {15} +gammasqrt{10}= x_3alpha+x_4betasqrt 6 \ alphasqrt{15}+betasqrt{10}+gammasqrt{6}=x_5alpha+x_6betasqrt 6end{cases}$$
which can be rewrite
$$MX=0quadtext{ with } quad M=begin{pmatrix} -x_1 & -x_2sqrt 6 & sqrt{15} \ sqrt{6}-x_3 & -sqrt{15}-x_4sqrt 6 & sqrt{10} \ sqrt{15}-x_5 & sqrt{10}- x_6sqrt 6 & sqrt 6 end{pmatrix}quad text{ and }quad X=begin{pmatrix} alpha \ beta\ gammaend{pmatrix}.$$
So if $det(M)ne 0$, we have won, since $(alpha,beta,gamma)=(0,0,0)$ is the only solution, thus $Y=0$ which is absurd.
Let's compute $det(M)$ then:
$$det(M)=Asqrt{15}+Bsqrt{10}+Csqrt 6+D,$$
with $A,B,C,Dinmathbb Q$ depending on the $x_i$. Since
$$dim_{mathbb Q}(mathrm{Span}_{mathbb Q}(sqrt 6,sqrt{10},sqrt{15}))=3,$$
and $A,B,C,Dinmathbb Q$, if we assume $det(M)=0$, we must have
$$A=B=C=D=0.$$
The computations give
$$(mathscr S)quad begin{cases} 2x_2x_5-2x_1x_6-6x_6+15=0 \ -3x_4x_5+3x_3x_6+3x_1=0 \ 6x_2-5x_3+15x_4=0 \ 6x_2x_3+6x_1x_4+10x_1-30x_2-15x_5+30=0.end{cases}$$
The question can now be reformulated as follow:
Does the system $(mathscr S)$ has rational solutions?
If we try to solve the system, we can end up with this expression:
$frac{20 , x_{1} x_{3}^{2} x_{5} + 12 , x_{1}^{3} - 30 , x_{1}^{2} x_{3} + 20 , x_{1}^{2} x_{5} + 30 , x_{3}^{2} x_{5} - 30 , x_{1} x_{5}^{2} - 75 , x_{3} x_{5}^{2} + 186 , x_{1}^{2} - 225 , x_{3}^{2} + 120 , x_{1} x_{5} - 90 , x_{5}^{2} + 450 , x_{1} + 1125 , x_{3} + 180 , x_{5}}{2 , x_{1} x_{5} + 5 , x_{3} x_{5} + 6 , x_{5}}=0.$
The question is then to understand if this surface in $mathbb R^3$ has rational points. If you are curious about it, this is what the numerator looks like:
Final remarks.
This question really interests me, but I feel quite stuck about it. May be my whole approach isn't going to help. Any hints, references or piece of solutions would be much appreciated.
linear-algebra number-theory vector-spaces diophantine-equations rational-numbers
asked Dec 1 '18 at 14:31
E. JosephE. Joseph
11.7k82856
$endgroup$
Some context.
By rational subspace, I mean a subspace of $mathbb R^5$ which admits a rational basis. In other words, a basis formed with vectors of $mathbb Q^5$.
For instance, the vector $v=(0,pi,3pi,pi-pi^2)$ is in a rational subspace of dimension $2$ since:
$$v=pibegin{pmatrix} 0 \ 1 \ 3 \ 1end{pmatrix}+pi^2begin{pmatrix} 0 \ 0 \ 0 \ -1end{pmatrix}.$$
The question.
Let $A=mathrm{Span}(Y_1,Y_2,Y_3)$ where
$$Y_1=begin{pmatrix} 1 \ 0 \ 0 \ sqrt 6 \ sqrt {15} end{pmatrix},quad Y_2=begin{pmatrix} 0 \ sqrt{6} \ 0 \ -sqrt {15} \ sqrt {10} end{pmatrix}quadtext { and }quad Y_3=begin{pmatrix} 0 \ 0 \ sqrt{15} \ sqrt {10} \ sqrt {6} end{pmatrix}.$$
Does there exist a rational subspace $B$ of $mathbb R^5$, of dimension $2$, such that $Acap Bne{0}$?
I believe the answer to be negative.
What I tried.
For $Yin A$, let's denote by $Y_1,ldots,Y_5inmathbb R$ its coordinates. We can prove the following lemma.
Lemma. The answer to the question is negative, if, and only if,
$$forall Y in A,quad dim_{mathbb Q}(mathrm{Span}_{mathbb Q}(Y_1,ldots,Y_5)ge 3.$$
I tried to investigate what such a rational subspace $B$ would look like. Let's take $B$, a rational subspace of $mathbb R^5$ of dimension $2$, such that $Acap Bne{0}$.
Let $Y:=alpha Y_1+beta Y_2+gamma Y_3$ be a vector in $Asetminus{0}$.
We have
$$alpha Y_1+beta Y_2+gamma Y_3=begin{pmatrix} alpha \ betasqrt 6 \ gamma sqrt {15} \ alphasqrt 6-betasqrt {15} +gammasqrt{10} \ alphasqrt{15}+betasqrt{10}+gammasqrt{6}end{pmatrix}.$$
If we assume (to try to get somewhere) that
$$dim_{mathbb Q}(mathrm{Span}_{mathbb Q}(alpha,sqrt{6}beta)=2,$$
then the last three coordinates of $Y$ must be a rational linear combination of the first two, i.e. there exists $x_1,ldots,x_6inmathbb Q$ such that
$$begin{cases}gamma sqrt {15}=x_1alpha+x_2betasqrt 6 \ alphasqrt 6-betasqrt {15} +gammasqrt{10}= x_3alpha+x_4betasqrt 6 \ alphasqrt{15}+betasqrt{10}+gammasqrt{6}=x_5alpha+x_6betasqrt 6end{cases}$$
which can be rewrite
$$MX=0quadtext{ with } quad M=begin{pmatrix} -x_1 & -x_2sqrt 6 & sqrt{15} \ sqrt{6}-x_3 & -sqrt{15}-x_4sqrt 6 & sqrt{10} \ sqrt{15}-x_5 & sqrt{10}- x_6sqrt 6 & sqrt 6 end{pmatrix}quad text{ and }quad X=begin{pmatrix} alpha \ beta\ gammaend{pmatrix}.$$
So if $det(M)ne 0$, we have won, since $(alpha,beta,gamma)=(0,0,0)$ is the only solution, thus $Y=0$ which is absurd.
Let's compute $det(M)$ then:
$$det(M)=Asqrt{15}+Bsqrt{10}+Csqrt 6+D,$$
with $A,B,C,Dinmathbb Q$ depending on the $x_i$. Since
$$dim_{mathbb Q}(mathrm{Span}_{mathbb Q}(sqrt 6,sqrt{10},sqrt{15}))=3,$$
and $A,B,C,Dinmathbb Q$, if we assume $det(M)=0$, we must have
$$A=B=C=D=0.$$
The computations give
$$(mathscr S)quad begin{cases} 2x_2x_5-2x_1x_6-6x_6+15=0 \ -3x_4x_5+3x_3x_6+3x_1=0 \ 6x_2-5x_3+15x_4=0 \ 6x_2x_3+6x_1x_4+10x_1-30x_2-15x_5+30=0.end{cases}$$
The question can now be reformulated as follow:
Does the system $(mathscr S)$ has rational solutions?
If we try to solve the system, we can end up with this expression:
$frac{20 , x_{1} x_{3}^{2} x_{5} + 12 , x_{1}^{3} - 30 , x_{1}^{2} x_{3} + 20 , x_{1}^{2} x_{5} + 30 , x_{3}^{2} x_{5} - 30 , x_{1} x_{5}^{2} - 75 , x_{3} x_{5}^{2} + 186 , x_{1}^{2} - 225 , x_{3}^{2} + 120 , x_{1} x_{5} - 90 , x_{5}^{2} + 450 , x_{1} + 1125 , x_{3} + 180 , x_{5}}{2 , x_{1} x_{5} + 5 , x_{3} x_{5} + 6 , x_{5}}=0.$
The question is then to understand if this surface in $mathbb R^3$ has rational points. If you are curious about it, this is what the numerator looks like:
Final remarks.
This question really interests me, but I feel quite stuck about it. May be my whole approach isn't going to help. Any hints, references or piece of solutions would be much appreciated.
linear-algebra number-theory vector-spaces diophantine-equations rational-numbers
linear-algebra number-theory vector-spaces diophantine-equations rational-numbers
asked Dec 1 '18 at 14:31
E. JosephE. Joseph
11.7k82856
asked Dec 1 '18 at 14:31
E. JosephE. Joseph
11.7k82856
edited Jan 13 at 9:25
E. Joseph
asked Dec 1 '18 at 14:31
E. JosephE. Joseph
11.7k82856
asked Dec 1 '18 at 14:31
E. JosephE. Joseph
11.7k82856
asked Dec 1 '18 at 14:31
E. JosephE. Joseph
11.7k82856
11.7k82856
1
$begingroup$
@AlexRavsky The lemma was indeed maybe confusing, so I made it a little more clear I hope.
$endgroup$
– E. Joseph
Jan 13 at 9:25
add a comment |
1
$begingroup$
@AlexRavsky The lemma was indeed maybe confusing, so I made it a little more clear I hope.
$endgroup$
– E. Joseph
Jan 13 at 9:25
1
1
$begingroup$
@AlexRavsky The lemma was indeed maybe confusing, so I made it a little more clear I hope.
$endgroup$
– E. Joseph
Jan 13 at 9:25
$begingroup$
@AlexRavsky The lemma was indeed maybe confusing, so I made it a little more clear I hope.
$endgroup$
– E. Joseph
Jan 13 at 9:25
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If your question boils down to the system,
$$(mathscr S)quad begin{cases} 2x_2x_5-2x_1x_6-6x_6+15=0 \ -3x_4x_5+3x_3x_6+3x_1=0 \ 6x_2-5x_3+15x_4=0 \ 6x_2x_3+6x_1x_4+10x_1-30x_2-15x_5+30=0.end{cases}$$
then, YES, this system has infinitely many rational solutions. If you let,
$$x_1 = x_4 x_5-x_3 x_6$$
$$x_2 = (5/6)(x_3-3x_4)$$
$$x_5 =frac{-3(15-6x_6+2x_3x_6^2)}{5x_3-15x_4-6x_4x_6}$$
This satisfies the first 3 equations and the 4th becomes a quadratic in $x_4$,
$$text{Poly}_1 x_4^2+text{Poly}_2 x_4+text{Poly}_3=0$$
You simply solve the linear equation,
$$text{Poly}_1 = -155 + 25 x_3 - 38 x_6 + 20 x_3 x_6 = 0$$
for $x_6$, thus,
$$x_4 =-frac{text{Poly}_3}{text{Poly}_2}$$
with free parameter $x_3$.
answered Dec 1 '18 at 15:28
Tito Piezas IIITito Piezas III
27.2k366174
$endgroup$
add a comment |
$begingroup$
Above simultaneous equations shown below has numerical solutions:
$(mathscr S)quad begin{cases} 2x_2x_5-2x_1x_6-6x_6+15=0 \ -3x_4x_5+3x_3x_6+3x_1=0 \ 6x_2-5x_3+15x_4=0 \ 6x_2x_3+6x_1x_4+10x_1-30x_2-15x_5+30=0.end{cases}$
$x_1$=$(-81255/10666)$
$x_2$=(32315/504)
$x_3$=$(2)$
$x_4$=$(-6295/252)$
$x_5$=$(-20790/5333)$
$x_6$=$(105/2)$
The solution provided by Tito Piezas in this context
actually requires solving a cubic equation instead of
a quadratic equation as mentioned by him.
(Note by Tito Piezas): My solution is a cubic in the variable $x_3$, but only a quadratic in $color{red}{x_4}$. This is easily verified using Mathematica.
edited Dec 30 '18 at 2:57
Tito Piezas III
27.2k366174
answered Dec 2 '18 at 15:40
SamSam
291
$endgroup$
$begingroup$
Thanks a lot, I really appreciate the explicit solution.
$endgroup$
– E. Joseph
Dec 2 '18 at 19:22
$begingroup$
You are welcome
$endgroup$
– Sam
Dec 2 '18 at 20:40
$begingroup$
In my answer, I was careful to say it is a quadratic in the variable $color{red}{x_4}$. Your comment applies to the variable $x_3$.
$endgroup$
– Tito Piezas III
Dec 30 '18 at 2:54
add a comment |
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2 Answers
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2 Answers
2
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$begingroup$
If your question boils down to the system,
$$(mathscr S)quad begin{cases} 2x_2x_5-2x_1x_6-6x_6+15=0 \ -3x_4x_5+3x_3x_6+3x_1=0 \ 6x_2-5x_3+15x_4=0 \ 6x_2x_3+6x_1x_4+10x_1-30x_2-15x_5+30=0.end{cases}$$
then, YES, this system has infinitely many rational solutions. If you let,
$$x_1 = x_4 x_5-x_3 x_6$$
$$x_2 = (5/6)(x_3-3x_4)$$
$$x_5 =frac{-3(15-6x_6+2x_3x_6^2)}{5x_3-15x_4-6x_4x_6}$$
This satisfies the first 3 equations and the 4th becomes a quadratic in $x_4$,
$$text{Poly}_1 x_4^2+text{Poly}_2 x_4+text{Poly}_3=0$$
You simply solve the linear equation,
$$text{Poly}_1 = -155 + 25 x_3 - 38 x_6 + 20 x_3 x_6 = 0$$
for $x_6$, thus,
$$x_4 =-frac{text{Poly}_3}{text{Poly}_2}$$
with free parameter $x_3$.
answered Dec 1 '18 at 15:28
Tito Piezas IIITito Piezas III
27.2k366174
$endgroup$
add a comment |
$begingroup$
If your question boils down to the system,
$$(mathscr S)quad begin{cases} 2x_2x_5-2x_1x_6-6x_6+15=0 \ -3x_4x_5+3x_3x_6+3x_1=0 \ 6x_2-5x_3+15x_4=0 \ 6x_2x_3+6x_1x_4+10x_1-30x_2-15x_5+30=0.end{cases}$$
then, YES, this system has infinitely many rational solutions. If you let,
$$x_1 = x_4 x_5-x_3 x_6$$
$$x_2 = (5/6)(x_3-3x_4)$$
$$x_5 =frac{-3(15-6x_6+2x_3x_6^2)}{5x_3-15x_4-6x_4x_6}$$
This satisfies the first 3 equations and the 4th becomes a quadratic in $x_4$,
$$text{Poly}_1 x_4^2+text{Poly}_2 x_4+text{Poly}_3=0$$
You simply solve the linear equation,
$$text{Poly}_1 = -155 + 25 x_3 - 38 x_6 + 20 x_3 x_6 = 0$$
for $x_6$, thus,
$$x_4 =-frac{text{Poly}_3}{text{Poly}_2}$$
with free parameter $x_3$.
answered Dec 1 '18 at 15:28
Tito Piezas IIITito Piezas III
27.2k366174
$endgroup$
add a comment |
$begingroup$
If your question boils down to the system,
$$(mathscr S)quad begin{cases} 2x_2x_5-2x_1x_6-6x_6+15=0 \ -3x_4x_5+3x_3x_6+3x_1=0 \ 6x_2-5x_3+15x_4=0 \ 6x_2x_3+6x_1x_4+10x_1-30x_2-15x_5+30=0.end{cases}$$
then, YES, this system has infinitely many rational solutions. If you let,
$$x_1 = x_4 x_5-x_3 x_6$$
$$x_2 = (5/6)(x_3-3x_4)$$
$$x_5 =frac{-3(15-6x_6+2x_3x_6^2)}{5x_3-15x_4-6x_4x_6}$$
This satisfies the first 3 equations and the 4th becomes a quadratic in $x_4$,
$$text{Poly}_1 x_4^2+text{Poly}_2 x_4+text{Poly}_3=0$$
You simply solve the linear equation,
$$text{Poly}_1 = -155 + 25 x_3 - 38 x_6 + 20 x_3 x_6 = 0$$
for $x_6$, thus,
$$x_4 =-frac{text{Poly}_3}{text{Poly}_2}$$
with free parameter $x_3$.
answered Dec 1 '18 at 15:28
Tito Piezas IIITito Piezas III
27.2k366174
$endgroup$
If your question boils down to the system,
$$(mathscr S)quad begin{cases} 2x_2x_5-2x_1x_6-6x_6+15=0 \ -3x_4x_5+3x_3x_6+3x_1=0 \ 6x_2-5x_3+15x_4=0 \ 6x_2x_3+6x_1x_4+10x_1-30x_2-15x_5+30=0.end{cases}$$
then, YES, this system has infinitely many rational solutions. If you let,
$$x_1 = x_4 x_5-x_3 x_6$$
$$x_2 = (5/6)(x_3-3x_4)$$
$$x_5 =frac{-3(15-6x_6+2x_3x_6^2)}{5x_3-15x_4-6x_4x_6}$$
This satisfies the first 3 equations and the 4th becomes a quadratic in $x_4$,
$$text{Poly}_1 x_4^2+text{Poly}_2 x_4+text{Poly}_3=0$$
You simply solve the linear equation,
$$text{Poly}_1 = -155 + 25 x_3 - 38 x_6 + 20 x_3 x_6 = 0$$
for $x_6$, thus,
$$x_4 =-frac{text{Poly}_3}{text{Poly}_2}$$
with free parameter $x_3$.
answered Dec 1 '18 at 15:28
Tito Piezas IIITito Piezas III
27.2k366174
answered Dec 1 '18 at 15:28
Tito Piezas IIITito Piezas III
27.2k366174
answered Dec 1 '18 at 15:28
Tito Piezas IIITito Piezas III
27.2k366174
answered Dec 1 '18 at 15:28
Tito Piezas IIITito Piezas III
27.2k366174
27.2k366174
add a comment |
add a comment |
$begingroup$
Above simultaneous equations shown below has numerical solutions:
$(mathscr S)quad begin{cases} 2x_2x_5-2x_1x_6-6x_6+15=0 \ -3x_4x_5+3x_3x_6+3x_1=0 \ 6x_2-5x_3+15x_4=0 \ 6x_2x_3+6x_1x_4+10x_1-30x_2-15x_5+30=0.end{cases}$
$x_1$=$(-81255/10666)$
$x_2$=(32315/504)
$x_3$=$(2)$
$x_4$=$(-6295/252)$
$x_5$=$(-20790/5333)$
$x_6$=$(105/2)$
The solution provided by Tito Piezas in this context
actually requires solving a cubic equation instead of
a quadratic equation as mentioned by him.
(Note by Tito Piezas): My solution is a cubic in the variable $x_3$, but only a quadratic in $color{red}{x_4}$. This is easily verified using Mathematica.
edited Dec 30 '18 at 2:57
Tito Piezas III
27.2k366174
answered Dec 2 '18 at 15:40
SamSam
291
$endgroup$
$begingroup$
Thanks a lot, I really appreciate the explicit solution.
$endgroup$
– E. Joseph
Dec 2 '18 at 19:22
$begingroup$
You are welcome
$endgroup$
– Sam
Dec 2 '18 at 20:40
$begingroup$
In my answer, I was careful to say it is a quadratic in the variable $color{red}{x_4}$. Your comment applies to the variable $x_3$.
$endgroup$
– Tito Piezas III
Dec 30 '18 at 2:54
add a comment |
$begingroup$
Above simultaneous equations shown below has numerical solutions:
$(mathscr S)quad begin{cases} 2x_2x_5-2x_1x_6-6x_6+15=0 \ -3x_4x_5+3x_3x_6+3x_1=0 \ 6x_2-5x_3+15x_4=0 \ 6x_2x_3+6x_1x_4+10x_1-30x_2-15x_5+30=0.end{cases}$
$x_1$=$(-81255/10666)$
$x_2$=(32315/504)
$x_3$=$(2)$
$x_4$=$(-6295/252)$
$x_5$=$(-20790/5333)$
$x_6$=$(105/2)$
The solution provided by Tito Piezas in this context
actually requires solving a cubic equation instead of
a quadratic equation as mentioned by him.
(Note by Tito Piezas): My solution is a cubic in the variable $x_3$, but only a quadratic in $color{red}{x_4}$. This is easily verified using Mathematica.
edited Dec 30 '18 at 2:57
Tito Piezas III
27.2k366174
answered Dec 2 '18 at 15:40
SamSam
291
$endgroup$
$begingroup$
Thanks a lot, I really appreciate the explicit solution.
$endgroup$
– E. Joseph
Dec 2 '18 at 19:22
$begingroup$
You are welcome
$endgroup$
– Sam
Dec 2 '18 at 20:40
$begingroup$
In my answer, I was careful to say it is a quadratic in the variable $color{red}{x_4}$. Your comment applies to the variable $x_3$.
$endgroup$
– Tito Piezas III
Dec 30 '18 at 2:54
add a comment |
$begingroup$
Above simultaneous equations shown below has numerical solutions:
$(mathscr S)quad begin{cases} 2x_2x_5-2x_1x_6-6x_6+15=0 \ -3x_4x_5+3x_3x_6+3x_1=0 \ 6x_2-5x_3+15x_4=0 \ 6x_2x_3+6x_1x_4+10x_1-30x_2-15x_5+30=0.end{cases}$
$x_1$=$(-81255/10666)$
$x_2$=(32315/504)
$x_3$=$(2)$
$x_4$=$(-6295/252)$
$x_5$=$(-20790/5333)$
$x_6$=$(105/2)$
The solution provided by Tito Piezas in this context
actually requires solving a cubic equation instead of
a quadratic equation as mentioned by him.
(Note by Tito Piezas): My solution is a cubic in the variable $x_3$, but only a quadratic in $color{red}{x_4}$. This is easily verified using Mathematica.
edited Dec 30 '18 at 2:57
Tito Piezas III
27.2k366174
answered Dec 2 '18 at 15:40
SamSam
291
$endgroup$
Above simultaneous equations shown below has numerical solutions:
$(mathscr S)quad begin{cases} 2x_2x_5-2x_1x_6-6x_6+15=0 \ -3x_4x_5+3x_3x_6+3x_1=0 \ 6x_2-5x_3+15x_4=0 \ 6x_2x_3+6x_1x_4+10x_1-30x_2-15x_5+30=0.end{cases}$
$x_1$=$(-81255/10666)$
$x_2$=(32315/504)
$x_3$=$(2)$
$x_4$=$(-6295/252)$
$x_5$=$(-20790/5333)$
$x_6$=$(105/2)$
The solution provided by Tito Piezas in this context
actually requires solving a cubic equation instead of
a quadratic equation as mentioned by him.
(Note by Tito Piezas): My solution is a cubic in the variable $x_3$, but only a quadratic in $color{red}{x_4}$. This is easily verified using Mathematica.
edited Dec 30 '18 at 2:57
Tito Piezas III
27.2k366174
answered Dec 2 '18 at 15:40
SamSam
291
edited Dec 30 '18 at 2:57
Tito Piezas III
27.2k366174
edited Dec 30 '18 at 2:57
Tito Piezas III
27.2k366174
edited Dec 30 '18 at 2:57
Tito Piezas III
27.2k366174
27.2k366174
answered Dec 2 '18 at 15:40
SamSam
291
answered Dec 2 '18 at 15:40
SamSam
291
answered Dec 2 '18 at 15:40
SamSam
291
291
$begingroup$
Thanks a lot, I really appreciate the explicit solution.
$endgroup$
– E. Joseph
Dec 2 '18 at 19:22
$begingroup$
You are welcome
$endgroup$
– Sam
Dec 2 '18 at 20:40
$begingroup$
In my answer, I was careful to say it is a quadratic in the variable $color{red}{x_4}$. Your comment applies to the variable $x_3$.
$endgroup$
– Tito Piezas III
Dec 30 '18 at 2:54
add a comment |
$begingroup$
Thanks a lot, I really appreciate the explicit solution.
$endgroup$
– E. Joseph
Dec 2 '18 at 19:22
$begingroup$
You are welcome
$endgroup$
– Sam
Dec 2 '18 at 20:40
$begingroup$
In my answer, I was careful to say it is a quadratic in the variable $color{red}{x_4}$. Your comment applies to the variable $x_3$.
$endgroup$
– Tito Piezas III
Dec 30 '18 at 2:54
$begingroup$
Thanks a lot, I really appreciate the explicit solution.
$endgroup$
– E. Joseph
Dec 2 '18 at 19:22
$begingroup$
Thanks a lot, I really appreciate the explicit solution.
$endgroup$
– E. Joseph
Dec 2 '18 at 19:22
$begingroup$
You are welcome
$endgroup$
– Sam
Dec 2 '18 at 20:40
$begingroup$
You are welcome
$endgroup$
– Sam
Dec 2 '18 at 20:40
$begingroup$
In my answer, I was careful to say it is a quadratic in the variable $color{red}{x_4}$. Your comment applies to the variable $x_3$.
$endgroup$
– Tito Piezas III
Dec 30 '18 at 2:54
$begingroup$
In my answer, I was careful to say it is a quadratic in the variable $color{red}{x_4}$. Your comment applies to the variable $x_3$.
$endgroup$
– Tito Piezas III
Dec 30 '18 at 2:54
add a comment |
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$begingroup$
@AlexRavsky The lemma was indeed maybe confusing, so I made it a little more clear I hope.
$endgroup$
– E. Joseph
Jan 13 at 9:25