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First of all every coefficient in alphabet letters in the equations be integers except $x$, $y$, $u$ and $v$,

I would like to know how a second-degree diophantine equation with two unknows like this,

$ax^2 + bxy + cy^2 + dx + ey + f = 0$

can be transformed using linear transformation into this, $mu^2 + nv^2 + o = 0$, where $u=rx+sy+t$ and $v=gx+hy+i$

For example, $9x^2 + 6xy - 13y^2 - 6x - 16y + 20 = 0$

There is : $2u^2 - 7v^2 + 45 = 0$, where $u = 3x + y - 1$ and $v = 2y + 1$

Someone told me that $mu^2 + nv^2 + o = 0$ could be found using the linear transformation method. How do we get to this equation using the linear transformation ?

Is this always possible ? And is there any other easier way?

Thank you !

Jotadiolyne Dicci
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1 Answers1

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$a (a x^2 + b x y + c y^2 + d x + e y + f)\implies$

$a^2 x^2 + x a(d + b y) + a f + a e y + a c y^2\implies$

$(a x + (d + b y)/2)^2 - (d + b y)^2/4 + a f + a e y + a c y^2\implies$

$(2 a x + b y + d)^2 - d^2 + 4 a f - 2 b d y + 4 a e y - b^2 y^2 + 4 a c y^2\implies$

$(2 a x + b y + d)^2 - d^2 + 4 a f + (4 a e-2 b d) y + (4 a c-b^2) y^2\implies$

$(4 a c-b^2) (2 a x + b y + d)^2 + (4 a c-b^2) (4 a f-d^2) + (4 a c-b^2) (4 a e-2 b d) y + (4 a c-b^2)^2 y^2\implies$

$(4 a c-b^2) (2 a x + b y + d)^2 + ((4 a c-b^2) y + (2 a e-b d))^2 - (2 a e-b d)^2 + (4 a c-b^2) (4 a f-d^2)$

Dmitry Ezhov
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