When do integers satisfy simple quadratic relation?












2












$begingroup$


$$((w+x)(y+z)-(w-x)(y-z))^2=4(wy+xz)^2-4(w^2-x^2)(y^2-z^2)$$ is true and so if $(a-b)^2=4(e+f)^2-4ab$ then under what additional minimal conditions can we say $e=wy$ and $f=xz$ while $a=(w+x)(y+z)$ and $b=(w-x)(y-z)$?



At least may be $a=(ell_1w+ell_2x)(ell_3y+ell_4z)$ and $b=(ell_1'w+ell_2'x)(ell_3'y+ell_4'z)$ holds at some special $ell_1,ell_2,ell_3,ell_4,ell_1',ell_2',ell_3',ell_4'inmathbb Z$? Does it seem possible?



From $$((w+x)(y+z)-(w-x)(y-z))^2=4(wy+xz)^2-4(w^2-x^2)(y^2-z^2)$$
with $w'=(w+x)$, $z'=(y+z)$, $x'=(w-x)$ and $y'=(y-z)$ we have
$$(w'z'-x'y')^2=(w'z'+x'y')^2-4cdot w'y'cdot x'z'$$
where
$$(w'y'+x'z')=((w+x)(y-z)+(w-x)(y+z))=(2wy+2xz)$$ holds.
We seek if $(a-b)^2=4(e+f)^2-4ab$ holds then is there a connection to $w',x',y',z'$ we can infer under some additional conditions?






number-theory polynomials algebraic-number-theory integers quadratic-forms







share|cite|improve this question











$endgroup$












  • $begingroup$
    It is not unambiguously clear to me what you are asking. Do you want to know when integers $a$, $b$, $e$, $f$ that satisfy $$(a-b)^2=4(e+f)^2-4ab,$$ are of the form $$e=wy,qquad f=xz,qquad a=(w+x)(y+z),qquad b=(w+x)(y+z),$$ for some integers $w$, $x$, $y$, $z$? Also, I think you mean $b=(w-x)(y-z)$?
    $endgroup$
    – Servaes
    Dec 3 '18 at 7:35






  • 1




    $begingroup$
    Related to an earlier, self-deleted question from OP, math.stackexchange.com/questions/3017854/…
    $endgroup$
    – Gerry Myerson
    Dec 3 '18 at 7:36










  • $begingroup$
    @Servaes $b=(w-x)(y-z)$.
    $endgroup$
    – Brout
    Dec 3 '18 at 7:37












  • $begingroup$
    @GerryMyerson That is not much use here I think.
    $endgroup$
    – Brout
    Dec 3 '18 at 7:38
















2












$begingroup$


$$((w+x)(y+z)-(w-x)(y-z))^2=4(wy+xz)^2-4(w^2-x^2)(y^2-z^2)$$ is true and so if $(a-b)^2=4(e+f)^2-4ab$ then under what additional minimal conditions can we say $e=wy$ and $f=xz$ while $a=(w+x)(y+z)$ and $b=(w-x)(y-z)$?



At least may be $a=(ell_1w+ell_2x)(ell_3y+ell_4z)$ and $b=(ell_1'w+ell_2'x)(ell_3'y+ell_4'z)$ holds at some special $ell_1,ell_2,ell_3,ell_4,ell_1',ell_2',ell_3',ell_4'inmathbb Z$? Does it seem possible?



From $$((w+x)(y+z)-(w-x)(y-z))^2=4(wy+xz)^2-4(w^2-x^2)(y^2-z^2)$$
with $w'=(w+x)$, $z'=(y+z)$, $x'=(w-x)$ and $y'=(y-z)$ we have
$$(w'z'-x'y')^2=(w'z'+x'y')^2-4cdot w'y'cdot x'z'$$
where
$$(w'y'+x'z')=((w+x)(y-z)+(w-x)(y+z))=(2wy+2xz)$$ holds.
We seek if $(a-b)^2=4(e+f)^2-4ab$ holds then is there a connection to $w',x',y',z'$ we can infer under some additional conditions?






number-theory polynomials algebraic-number-theory integers quadratic-forms







share|cite|improve this question











$endgroup$












  • $begingroup$
    It is not unambiguously clear to me what you are asking. Do you want to know when integers $a$, $b$, $e$, $f$ that satisfy $$(a-b)^2=4(e+f)^2-4ab,$$ are of the form $$e=wy,qquad f=xz,qquad a=(w+x)(y+z),qquad b=(w+x)(y+z),$$ for some integers $w$, $x$, $y$, $z$? Also, I think you mean $b=(w-x)(y-z)$?
    $endgroup$
    – Servaes
    Dec 3 '18 at 7:35






  • 1




    $begingroup$
    Related to an earlier, self-deleted question from OP, math.stackexchange.com/questions/3017854/…
    $endgroup$
    – Gerry Myerson
    Dec 3 '18 at 7:36










  • $begingroup$
    @Servaes $b=(w-x)(y-z)$.
    $endgroup$
    – Brout
    Dec 3 '18 at 7:37












  • $begingroup$
    @GerryMyerson That is not much use here I think.
    $endgroup$
    – Brout
    Dec 3 '18 at 7:38














2












2








2





$begingroup$


$$((w+x)(y+z)-(w-x)(y-z))^2=4(wy+xz)^2-4(w^2-x^2)(y^2-z^2)$$ is true and so if $(a-b)^2=4(e+f)^2-4ab$ then under what additional minimal conditions can we say $e=wy$ and $f=xz$ while $a=(w+x)(y+z)$ and $b=(w-x)(y-z)$?



At least may be $a=(ell_1w+ell_2x)(ell_3y+ell_4z)$ and $b=(ell_1'w+ell_2'x)(ell_3'y+ell_4'z)$ holds at some special $ell_1,ell_2,ell_3,ell_4,ell_1',ell_2',ell_3',ell_4'inmathbb Z$? Does it seem possible?



From $$((w+x)(y+z)-(w-x)(y-z))^2=4(wy+xz)^2-4(w^2-x^2)(y^2-z^2)$$
with $w'=(w+x)$, $z'=(y+z)$, $x'=(w-x)$ and $y'=(y-z)$ we have
$$(w'z'-x'y')^2=(w'z'+x'y')^2-4cdot w'y'cdot x'z'$$
where
$$(w'y'+x'z')=((w+x)(y-z)+(w-x)(y+z))=(2wy+2xz)$$ holds.
We seek if $(a-b)^2=4(e+f)^2-4ab$ holds then is there a connection to $w',x',y',z'$ we can infer under some additional conditions?






number-theory polynomials algebraic-number-theory integers quadratic-forms







share|cite|improve this question











$endgroup$




$$((w+x)(y+z)-(w-x)(y-z))^2=4(wy+xz)^2-4(w^2-x^2)(y^2-z^2)$$ is true and so if $(a-b)^2=4(e+f)^2-4ab$ then under what additional minimal conditions can we say $e=wy$ and $f=xz$ while $a=(w+x)(y+z)$ and $b=(w-x)(y-z)$?



At least may be $a=(ell_1w+ell_2x)(ell_3y+ell_4z)$ and $b=(ell_1'w+ell_2'x)(ell_3'y+ell_4'z)$ holds at some special $ell_1,ell_2,ell_3,ell_4,ell_1',ell_2',ell_3',ell_4'inmathbb Z$? Does it seem possible?



From $$((w+x)(y+z)-(w-x)(y-z))^2=4(wy+xz)^2-4(w^2-x^2)(y^2-z^2)$$
with $w'=(w+x)$, $z'=(y+z)$, $x'=(w-x)$ and $y'=(y-z)$ we have
$$(w'z'-x'y')^2=(w'z'+x'y')^2-4cdot w'y'cdot x'z'$$
where
$$(w'y'+x'z')=((w+x)(y-z)+(w-x)(y+z))=(2wy+2xz)$$ holds.
We seek if $(a-b)^2=4(e+f)^2-4ab$ holds then is there a connection to $w',x',y',z'$ we can infer under some additional conditions?






number-theory polynomials algebraic-number-theory integers quadratic-forms




number-theory polynomials algebraic-number-theory integers quadratic-forms






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 3 '18 at 7:37







Brout

















asked Dec 3 '18 at 6:50









BroutBrout

2,1571428




2,1571428












  • $begingroup$
    It is not unambiguously clear to me what you are asking. Do you want to know when integers $a$, $b$, $e$, $f$ that satisfy $$(a-b)^2=4(e+f)^2-4ab,$$ are of the form $$e=wy,qquad f=xz,qquad a=(w+x)(y+z),qquad b=(w+x)(y+z),$$ for some integers $w$, $x$, $y$, $z$? Also, I think you mean $b=(w-x)(y-z)$?
    $endgroup$
    – Servaes
    Dec 3 '18 at 7:35






  • 1




    $begingroup$
    Related to an earlier, self-deleted question from OP, math.stackexchange.com/questions/3017854/…
    $endgroup$
    – Gerry Myerson
    Dec 3 '18 at 7:36










  • $begingroup$
    @Servaes $b=(w-x)(y-z)$.
    $endgroup$
    – Brout
    Dec 3 '18 at 7:37












  • $begingroup$
    @GerryMyerson That is not much use here I think.
    $endgroup$
    – Brout
    Dec 3 '18 at 7:38


















  • $begingroup$
    It is not unambiguously clear to me what you are asking. Do you want to know when integers $a$, $b$, $e$, $f$ that satisfy $$(a-b)^2=4(e+f)^2-4ab,$$ are of the form $$e=wy,qquad f=xz,qquad a=(w+x)(y+z),qquad b=(w+x)(y+z),$$ for some integers $w$, $x$, $y$, $z$? Also, I think you mean $b=(w-x)(y-z)$?
    $endgroup$
    – Servaes
    Dec 3 '18 at 7:35






  • 1




    $begingroup$
    Related to an earlier, self-deleted question from OP, math.stackexchange.com/questions/3017854/…
    $endgroup$
    – Gerry Myerson
    Dec 3 '18 at 7:36










  • $begingroup$
    @Servaes $b=(w-x)(y-z)$.
    $endgroup$
    – Brout
    Dec 3 '18 at 7:37












  • $begingroup$
    @GerryMyerson That is not much use here I think.
    $endgroup$
    – Brout
    Dec 3 '18 at 7:38
















$begingroup$
It is not unambiguously clear to me what you are asking. Do you want to know when integers $a$, $b$, $e$, $f$ that satisfy $$(a-b)^2=4(e+f)^2-4ab,$$ are of the form $$e=wy,qquad f=xz,qquad a=(w+x)(y+z),qquad b=(w+x)(y+z),$$ for some integers $w$, $x$, $y$, $z$? Also, I think you mean $b=(w-x)(y-z)$?
$endgroup$
– Servaes
Dec 3 '18 at 7:35




$begingroup$
It is not unambiguously clear to me what you are asking. Do you want to know when integers $a$, $b$, $e$, $f$ that satisfy $$(a-b)^2=4(e+f)^2-4ab,$$ are of the form $$e=wy,qquad f=xz,qquad a=(w+x)(y+z),qquad b=(w+x)(y+z),$$ for some integers $w$, $x$, $y$, $z$? Also, I think you mean $b=(w-x)(y-z)$?
$endgroup$
– Servaes
Dec 3 '18 at 7:35




1




1




$begingroup$
Related to an earlier, self-deleted question from OP, math.stackexchange.com/questions/3017854/…
$endgroup$
– Gerry Myerson
Dec 3 '18 at 7:36




$begingroup$
Related to an earlier, self-deleted question from OP, math.stackexchange.com/questions/3017854/…
$endgroup$
– Gerry Myerson
Dec 3 '18 at 7:36












$begingroup$
@Servaes $b=(w-x)(y-z)$.
$endgroup$
– Brout
Dec 3 '18 at 7:37






$begingroup$
@Servaes $b=(w-x)(y-z)$.
$endgroup$
– Brout
Dec 3 '18 at 7:37














$begingroup$
@GerryMyerson That is not much use here I think.
$endgroup$
– Brout
Dec 3 '18 at 7:38




$begingroup$
@GerryMyerson That is not much use here I think.
$endgroup$
– Brout
Dec 3 '18 at 7:38










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