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:

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


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

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

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

for arbitrary $p,q$.

Identity 2:

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

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


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

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

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

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

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 $(5)$?

See also this post.

  • $\begingroup$ And what's the point of all this? This is a dead-end direction. Because such transformations, the equation cannot be solved. In fact, replacing one equation by the other. Actually solving the binary form is easy to come to the Pell equation. I showed there. math.stackexchange.com/questions/580491/… We need to think of ways to simplify the calculations, not to complicate things. $\endgroup$ – individ Nov 29 '14 at 5:01
  • 1
    $\begingroup$ 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. $\endgroup$ – Tito Piezas III Nov 29 '14 at 9:06
  • $\begingroup$ People are very conservative. Very hard to explain that this is the same approach. I have the above formula is easier to write the same equations. In a more compact form of the formula look like this. I prefer the more intuitive formula. $\endgroup$ – individ Nov 29 '14 at 9:20

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