Does this polynomial have a rational value which is the square of a rational number?
up vote
4
down vote
favorite
I have the following polynomial:
$$P(x,y,z):=9y^2z^2-30x^2z+90xyz+54yz-270x+81inmathbb Q[x].$$
It came up in a larger proof, and I would need in order to complete the proof to prove the following result:
Does there exist $(x,y,z,r)inmathbb Q^4$ such that $xne 0$ and
$$P(x,y,z)=r^2.$$
We can reformulate the problem in the following way:
Does the algebraic variety defined by
$$9Y^2Z^2-30X^2Z+90XYZ+54YZ-270X+81-T^2$$
have a rational point with $Xne 0$?
I have no idea how to tackle this problem, I have looked up several articles, but nothing seems to apply to this particular question.
Any hints or references would be greatly appreciated.
number-theory algebraic-geometry polynomials diophantine-equations rational-numbers
edited Nov 29 at 11:12
Servaes
21.5k33792
asked Nov 29 at 10:47
E. Joseph
11.5k82856
add a comment |
up vote
4
down vote
favorite
I have the following polynomial:
$$P(x,y,z):=9y^2z^2-30x^2z+90xyz+54yz-270x+81inmathbb Q[x].$$
It came up in a larger proof, and I would need in order to complete the proof to prove the following result:
Does there exist $(x,y,z,r)inmathbb Q^4$ such that $xne 0$ and
$$P(x,y,z)=r^2.$$
We can reformulate the problem in the following way:
Does the algebraic variety defined by
$$9Y^2Z^2-30X^2Z+90XYZ+54YZ-270X+81-T^2$$
have a rational point with $Xne 0$?
I have no idea how to tackle this problem, I have looked up several articles, but nothing seems to apply to this particular question.
Any hints or references would be greatly appreciated.
number-theory algebraic-geometry polynomials diophantine-equations rational-numbers
edited Nov 29 at 11:12
Servaes
21.5k33792
asked Nov 29 at 10:47
E. Joseph
11.5k82856
@TonyK Yes, indeed, I have edited (I forgot I am not allowed to take $x=0$), sorry for the inconvenience.
– E. Joseph
Nov 29 at 10:54
add a comment |
up vote
4
down vote
favorite
up vote
4
down vote
favorite
I have the following polynomial:
$$P(x,y,z):=9y^2z^2-30x^2z+90xyz+54yz-270x+81inmathbb Q[x].$$
It came up in a larger proof, and I would need in order to complete the proof to prove the following result:
Does there exist $(x,y,z,r)inmathbb Q^4$ such that $xne 0$ and
$$P(x,y,z)=r^2.$$
We can reformulate the problem in the following way:
Does the algebraic variety defined by
$$9Y^2Z^2-30X^2Z+90XYZ+54YZ-270X+81-T^2$$
have a rational point with $Xne 0$?
I have no idea how to tackle this problem, I have looked up several articles, but nothing seems to apply to this particular question.
Any hints or references would be greatly appreciated.
number-theory algebraic-geometry polynomials diophantine-equations rational-numbers
edited Nov 29 at 11:12
Servaes
21.5k33792
asked Nov 29 at 10:47
E. Joseph
11.5k82856
I have the following polynomial:
$$P(x,y,z):=9y^2z^2-30x^2z+90xyz+54yz-270x+81inmathbb Q[x].$$
It came up in a larger proof, and I would need in order to complete the proof to prove the following result:
Does there exist $(x,y,z,r)inmathbb Q^4$ such that $xne 0$ and
$$P(x,y,z)=r^2.$$
We can reformulate the problem in the following way:
Does the algebraic variety defined by
$$9Y^2Z^2-30X^2Z+90XYZ+54YZ-270X+81-T^2$$
have a rational point with $Xne 0$?
I have no idea how to tackle this problem, I have looked up several articles, but nothing seems to apply to this particular question.
Any hints or references would be greatly appreciated.
number-theory algebraic-geometry polynomials diophantine-equations rational-numbers
number-theory algebraic-geometry polynomials diophantine-equations rational-numbers
edited Nov 29 at 11:12
Servaes
21.5k33792
asked Nov 29 at 10:47
E. Joseph
11.5k82856
edited Nov 29 at 11:12
Servaes
21.5k33792
asked Nov 29 at 10:47
E. Joseph
11.5k82856
edited Nov 29 at 11:12
Servaes
21.5k33792
edited Nov 29 at 11:12
Servaes
21.5k33792
edited Nov 29 at 11:12
Servaes
21.5k33792
21.5k33792
asked Nov 29 at 10:47
E. Joseph
11.5k82856
asked Nov 29 at 10:47
E. Joseph
11.5k82856
asked Nov 29 at 10:47
E. Joseph
11.5k82856
11.5k82856
@TonyK Yes, indeed, I have edited (I forgot I am not allowed to take $x=0$), sorry for the inconvenience.
– E. Joseph
Nov 29 at 10:54
add a comment |
@TonyK Yes, indeed, I have edited (I forgot I am not allowed to take $x=0$), sorry for the inconvenience.
– E. Joseph
Nov 29 at 10:54
@TonyK Yes, indeed, I have edited (I forgot I am not allowed to take $x=0$), sorry for the inconvenience.
– E. Joseph
Nov 29 at 10:54
@TonyK Yes, indeed, I have edited (I forgot I am not allowed to take $x=0$), sorry for the inconvenience.
– E. Joseph
Nov 29 at 10:54
add a comment |
1 Answer
1
active
oldest
votes
up vote
7
down vote
accepted
Two obvious solutions are $P(0,0,0)=(pm9)^2$.
To find more solutions, plugging in $z=0$ yields
$$P(x,y,0)=-270x+81,$$
which is a square for $x=frac{81-t^2}{270}$ for any choice of $tinBbb{Q}$, and any choice of $yinBbb{Q}$.
answered Nov 29 at 10:49
Servaes
21.5k33792
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
7
down vote
accepted
Two obvious solutions are $P(0,0,0)=(pm9)^2$.
To find more solutions, plugging in $z=0$ yields
$$P(x,y,0)=-270x+81,$$
which is a square for $x=frac{81-t^2}{270}$ for any choice of $tinBbb{Q}$, and any choice of $yinBbb{Q}$.
answered Nov 29 at 10:49
Servaes
21.5k33792
add a comment |
up vote
7
down vote
accepted
Two obvious solutions are $P(0,0,0)=(pm9)^2$.
To find more solutions, plugging in $z=0$ yields
$$P(x,y,0)=-270x+81,$$
which is a square for $x=frac{81-t^2}{270}$ for any choice of $tinBbb{Q}$, and any choice of $yinBbb{Q}$.
answered Nov 29 at 10:49
Servaes
21.5k33792
add a comment |
up vote
7
down vote
accepted
up vote
7
down vote
accepted
Two obvious solutions are $P(0,0,0)=(pm9)^2$.
To find more solutions, plugging in $z=0$ yields
$$P(x,y,0)=-270x+81,$$
which is a square for $x=frac{81-t^2}{270}$ for any choice of $tinBbb{Q}$, and any choice of $yinBbb{Q}$.
answered Nov 29 at 10:49
Servaes
21.5k33792
Two obvious solutions are $P(0,0,0)=(pm9)^2$.
To find more solutions, plugging in $z=0$ yields
$$P(x,y,0)=-270x+81,$$
which is a square for $x=frac{81-t^2}{270}$ for any choice of $tinBbb{Q}$, and any choice of $yinBbb{Q}$.
answered Nov 29 at 10:49
Servaes
21.5k33792
edited Nov 29 at 11:10
answered Nov 29 at 10:49
Servaes
21.5k33792
answered Nov 29 at 10:49
Servaes
21.5k33792
answered Nov 29 at 10:49
Servaes
21.5k33792
21.5k33792
add a comment |
add a comment |
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@TonyK Yes, indeed, I have edited (I forgot I am not allowed to take $x=0$), sorry for the inconvenience.
– E. Joseph
Nov 29 at 10:54