Quadratic Invariant

Given the Binary Quadratic Form

$$ax^2+2bxy+cy^2$$ (1)

with Discriminant $b^2-ac$, let

$$x = pX+qY$$ (2)

$$y = rX+sY.$$ (3)

Then

$$a(pX+qY)^2+2b(pX+qY)(rX+sY)+c(rX+sY)^2=AX^2+2BXY+CY^2,$$ (4)

where

$$A = ap^2+2bpr+cr^2$$ (5)

$$B = apq+b(ps+qr)+crs$$ (6)

$$C = aq^2+2bqs+cs^2,$$ (7)

so

$B^2-AC=[a^2p^2q^2+b^2(ps+qr)^2+c^2r^2s^2$
$\quad\phantom{=} +2abpq(ps+qr)+2acpqrs+2bcrs(ps+qr)]$
$\quad\phantom{=} -(ap^2+2bpr+cr^2)(aq^2+2bqs+cs^2)$
$\quad = a^2p^2q^2+b^2p^2s^2+2b^2pqrs+b^2q^2r^2+c^2r^2s^2$
$\quad\phantom{=} +2abp^2qs+2abpq^2r+2acpqrs+2bcprs^2+2bcqr^2s$
$\quad\phantom{=} -a^2p^2q^2-2abp^2qs-acp^2s^2-2abpq^2r-4b^2pqrs$
$\quad\phantom{=} -2bcprs^2-acq^2r^2-2bcqr^2s-c^2r^2s^2$
$\quad = b^2p^2s^2-2b^2pqrs+b^2q^2r^2+2acpqrs-acp^2s^2$
$\quad\phantom{=} -acq^2r^2$
$\quad = p^2s^2(b^2-ac)+q^2r^2(b^2-ac)-2pqrs(b^2-ac)$
$\quad = (b^2-ac)(p^2s^2-2pqrs+q^2r^2)$
$\quad = (ps-rq)^2(b^2-ac).$	(8)

Surprisingly, this is the same discriminant as before, but multiplied by the factor $(ps-rq)^2$. The quantity $ps-rq$ is called the Modulus.

See also Algebraic Invariant