A binary quadratic form whose range topograph has a lake
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Is it true that if Q is a binary quadratic form whose range topograph has a lake, then Q factors as a product of two linear forms with integer coefficients?
number-theory quadratic-forms
asked Dec 11 '18 at 7:32
Jingting931015Jingting931015
828
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add a comment |
$begingroup$
Is it true that if Q is a binary quadratic form whose range topograph has a lake, then Q factors as a product of two linear forms with integer coefficients?
number-theory quadratic-forms
asked Dec 11 '18 at 7:32
Jingting931015Jingting931015
828
$endgroup$
add a comment |
$begingroup$
Is it true that if Q is a binary quadratic form whose range topograph has a lake, then Q factors as a product of two linear forms with integer coefficients?
number-theory quadratic-forms
asked Dec 11 '18 at 7:32
Jingting931015Jingting931015
828
$endgroup$
Is it true that if Q is a binary quadratic form whose range topograph has a lake, then Q factors as a product of two linear forms with integer coefficients?
number-theory quadratic-forms
number-theory quadratic-forms
asked Dec 11 '18 at 7:32
Jingting931015Jingting931015
828
asked Dec 11 '18 at 7:32
Jingting931015Jingting931015
828
asked Dec 11 '18 at 7:32
Jingting931015Jingting931015
828
asked Dec 11 '18 at 7:32
Jingting931015Jingting931015
828
asked Dec 11 '18 at 7:32
Jingting931015Jingting931015
828
828
add a comment |
add a comment |
1 Answer
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Alright, Weissman's book, which I have. Very nice.
On page 274, definition 10.21 says a lake is a place where the value of the form is zero. I don't see it, but any vector in the "domain topograph" is nonzero, at least one of the elements of $(x,y)$ is nonzero. This follows from the fact that we extend from the "home" basis. Any lax basis, a pair of vectors with $pm$ signs, gives determinant $pm 1$ for the little 2 by 2 matrix. See note upper right on page 233.
That is about all we need. A lake says we have integers $u,v$ not both zero, with
$$ au^2 + b uv + c v^2 = 0 ; . ; $$
Now, if $v=0,$ we know that $u neq 0,$ so $a=0.$ In this special case, we can factor $bxy + c y^2$ as $(bx+cy)y.$
If $v neq 0,$ we can divide through by $v^2, $ then define rational $r = frac{u}{v},$ so that
$$ a r^2 + b r + c = 0 . $$
As we have a rational root, the Quadratic Formula says that the traditional discriminant
$$ Delta = b^2 - 4ac $$
must be an integer square, $Delta = delta^2.$ Then we can factor $ax^2 + bxy + c y^2$ over the integers, I give a method and proof at Prove that if $b^2-4ac=k^2$ then $ax^2+bx+c$ is factorizable
answered Dec 11 '18 at 19:05
Will JagyWill Jagy
103k5101200
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$begingroup$
Alright, Weissman's book, which I have. Very nice.
On page 274, definition 10.21 says a lake is a place where the value of the form is zero. I don't see it, but any vector in the "domain topograph" is nonzero, at least one of the elements of $(x,y)$ is nonzero. This follows from the fact that we extend from the "home" basis. Any lax basis, a pair of vectors with $pm$ signs, gives determinant $pm 1$ for the little 2 by 2 matrix. See note upper right on page 233.
That is about all we need. A lake says we have integers $u,v$ not both zero, with
$$ au^2 + b uv + c v^2 = 0 ; . ; $$
Now, if $v=0,$ we know that $u neq 0,$ so $a=0.$ In this special case, we can factor $bxy + c y^2$ as $(bx+cy)y.$
If $v neq 0,$ we can divide through by $v^2, $ then define rational $r = frac{u}{v},$ so that
$$ a r^2 + b r + c = 0 . $$
As we have a rational root, the Quadratic Formula says that the traditional discriminant
$$ Delta = b^2 - 4ac $$
must be an integer square, $Delta = delta^2.$ Then we can factor $ax^2 + bxy + c y^2$ over the integers, I give a method and proof at Prove that if $b^2-4ac=k^2$ then $ax^2+bx+c$ is factorizable
answered Dec 11 '18 at 19:05
Will JagyWill Jagy
103k5101200
$endgroup$
add a comment |
$begingroup$
Alright, Weissman's book, which I have. Very nice.
On page 274, definition 10.21 says a lake is a place where the value of the form is zero. I don't see it, but any vector in the "domain topograph" is nonzero, at least one of the elements of $(x,y)$ is nonzero. This follows from the fact that we extend from the "home" basis. Any lax basis, a pair of vectors with $pm$ signs, gives determinant $pm 1$ for the little 2 by 2 matrix. See note upper right on page 233.
That is about all we need. A lake says we have integers $u,v$ not both zero, with
$$ au^2 + b uv + c v^2 = 0 ; . ; $$
Now, if $v=0,$ we know that $u neq 0,$ so $a=0.$ In this special case, we can factor $bxy + c y^2$ as $(bx+cy)y.$
If $v neq 0,$ we can divide through by $v^2, $ then define rational $r = frac{u}{v},$ so that
$$ a r^2 + b r + c = 0 . $$
As we have a rational root, the Quadratic Formula says that the traditional discriminant
$$ Delta = b^2 - 4ac $$
must be an integer square, $Delta = delta^2.$ Then we can factor $ax^2 + bxy + c y^2$ over the integers, I give a method and proof at Prove that if $b^2-4ac=k^2$ then $ax^2+bx+c$ is factorizable
answered Dec 11 '18 at 19:05
Will JagyWill Jagy
103k5101200
$endgroup$
add a comment |
$begingroup$
Alright, Weissman's book, which I have. Very nice.
On page 274, definition 10.21 says a lake is a place where the value of the form is zero. I don't see it, but any vector in the "domain topograph" is nonzero, at least one of the elements of $(x,y)$ is nonzero. This follows from the fact that we extend from the "home" basis. Any lax basis, a pair of vectors with $pm$ signs, gives determinant $pm 1$ for the little 2 by 2 matrix. See note upper right on page 233.
That is about all we need. A lake says we have integers $u,v$ not both zero, with
$$ au^2 + b uv + c v^2 = 0 ; . ; $$
Now, if $v=0,$ we know that $u neq 0,$ so $a=0.$ In this special case, we can factor $bxy + c y^2$ as $(bx+cy)y.$
If $v neq 0,$ we can divide through by $v^2, $ then define rational $r = frac{u}{v},$ so that
$$ a r^2 + b r + c = 0 . $$
As we have a rational root, the Quadratic Formula says that the traditional discriminant
$$ Delta = b^2 - 4ac $$
must be an integer square, $Delta = delta^2.$ Then we can factor $ax^2 + bxy + c y^2$ over the integers, I give a method and proof at Prove that if $b^2-4ac=k^2$ then $ax^2+bx+c$ is factorizable
answered Dec 11 '18 at 19:05
Will JagyWill Jagy
103k5101200
$endgroup$
Alright, Weissman's book, which I have. Very nice.
On page 274, definition 10.21 says a lake is a place where the value of the form is zero. I don't see it, but any vector in the "domain topograph" is nonzero, at least one of the elements of $(x,y)$ is nonzero. This follows from the fact that we extend from the "home" basis. Any lax basis, a pair of vectors with $pm$ signs, gives determinant $pm 1$ for the little 2 by 2 matrix. See note upper right on page 233.
That is about all we need. A lake says we have integers $u,v$ not both zero, with
$$ au^2 + b uv + c v^2 = 0 ; . ; $$
Now, if $v=0,$ we know that $u neq 0,$ so $a=0.$ In this special case, we can factor $bxy + c y^2$ as $(bx+cy)y.$
If $v neq 0,$ we can divide through by $v^2, $ then define rational $r = frac{u}{v},$ so that
$$ a r^2 + b r + c = 0 . $$
As we have a rational root, the Quadratic Formula says that the traditional discriminant
$$ Delta = b^2 - 4ac $$
must be an integer square, $Delta = delta^2.$ Then we can factor $ax^2 + bxy + c y^2$ over the integers, I give a method and proof at Prove that if $b^2-4ac=k^2$ then $ax^2+bx+c$ is factorizable
answered Dec 11 '18 at 19:05
Will JagyWill Jagy
103k5101200
answered Dec 11 '18 at 19:05
Will JagyWill Jagy
103k5101200
answered Dec 11 '18 at 19:05
Will JagyWill Jagy
103k5101200
answered Dec 11 '18 at 19:05
Will JagyWill Jagy
103k5101200
103k5101200
add a comment |
add a comment |
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