# A Tale of Two Quadratic Identities (Pell-like)

Question is at the end. Let all variables be integers. For some constants $a,b,c,d$, assume we have initial solution {$m,n$} to,

$$a m^2 + b m n + c n^2 = d\tag{1}$$

Identity 1:

$$a x^2 + b x y + c y^2 - d z^2 = (a m^2 + b m n + c n^2 - d)(a p^2+bpq+c q^2)^2\tag{2}$$

where,

$$x =(am+bn)p^2+2cnpq-cmq^2$$

$$y = -anp^2+2ampq+(bm+cn)q^2$$

$$z = z_1 = ap^2+bpq+cq^2$$

for arbitrary $p,q$.

Identity 2:

Given initial solutions $m,n$ to $(1)$. Then,

$$a x^2 + b x y + c y^2 - d z^2 = (a m^2 + b m n + c n^2 - d)(x/m)^2\tag{3}$$

where,

$$x = mu^2 - 2(b m + 2c n)u v + D m v^2$$

$$y = n u^2 + 2(2a m + b n)u v + D n v^2$$

$$z = z_2 = u^2 - D v^2$$

$$D = b^2-4ac$$

for arbitrary $u,v$.

Questions: Let,

$$z_1 = ap^2+bpq+cq^2 =\pm 1\tag{4}$$

$$z_2 = u^2 - D v^2 = \pm 1\tag{5}$$

Is it true:

1. There are times we can solve $(4)$ in the integers, but not $(5)$?
2. Likewise, there are times we can solve $(5)$, but not $(4)$?
3. Or does being able to solve $(4)$ imply solvability of $(5)$?

• The question of whether $am^2 +bmn+cn^2 =d$ has an infinite number of integer solutions is reduced to solving $(4)$ or $(6)$, namely $ap^2 +bpq+cq^2 =\pm1$ or $u^2-Dv^2 = \pm1$, and aspects of it are discussed in Dickson's classic History of the Theory of Numbers. Most people, myself included, subscribe to Dickson's approach. – Tito Piezas III Nov 29 '14 at 9:06