4
$\begingroup$

In Carmichael's Diophantine Analysis ($\S8$), he notes that the equation $$X^2-dY^2=Z^2 \qquad(\dagger)$$ has a two-parameter solution $$x=m^2+dn^2, \quad y=2mn, \quad z=m^2-dn^2. \qquad(\star)$$ He then says “there is no ready means for determining whether this is the general solution”, and goes on to give a [much more complicated] general solution.

Is Carmichael’s conclusion true? i.e. Can it neither be proven nor disproven [easily] that $(\star)$ is indeed a general parametric solution to $(\dagger)$?

EDIT: In Barbeau’s Pell’s Equation (pg. 36), it says:

Exercise 2.8. [...] (a) Consider the case d=2 [of the equation $U^2-W^2=dV^2$]. Obtain the parametric solutions $$(u,v,w)=(r^2+2s^2,2rs,2s^2-r^2)$$ and $$(u,v,w)=(2r^2+s^2,2rs,s^2-2r^2).$$ (b) Obtain a parametric set of solutions for $u^2-dv^2=w^2$.

Given that this ‘exercise’ appears early in such an elementary textbook, it seems likely to me that there is a solution similar to $(\star)$ which is valid [though clearly $(\star)$ is not, as pointed out in the comments].

EDIT: Maybe Barbeau means let $d=d_1d_2$ and $v=v_1v_2$ be arbitrary factorizations, so that $dv^2=u^2-w^2=(u-w)(u+w)$ implies $u-w=d_1v_1^2$ and $u+w=d_2v_2^2$, yielding the parametric solution $$(u,v,w) = \biggl(\frac{d_2v_2^2+d_1v_1^2}{2},v_1v_2,\frac{d_2v_2^2-d_1v_1^2}{2}\biggr).$$

$\endgroup$
  • 1
    $\begingroup$ Obviously $y$ from $(\star)\,$is always even, so the solution $x=2, y=1$ for $d=3, z=1$ cannot be produced. $\endgroup$ – gammatester Jun 26 '14 at 13:25
  • $\begingroup$ Having just read the section, I'm not sure Carmichael is saying that it is unknown, but rather that the previous proof technique used before gives us no way to assert this. As previous commenter noted, this will only determine the points up to multiples. $\endgroup$ – Thomas Andrews Jun 26 '14 at 13:54
  • $\begingroup$ This formula really makes all the decisions. The formula has a General view there. math.stackexchange.com/questions/738446/… Interestingly, the formula for the equation $x^2+y^2=dz^2$ so simply cannot be recorded. It has a bulky look $\endgroup$ – individ Jun 26 '14 at 14:03
  • 1
    $\begingroup$ Just for reference of others, here's a link to the book, which is public domain: gutenberg.org/files/20073/20073-pdf.pdf $\endgroup$ – Thomas Andrews Jun 26 '14 at 14:07
  • 1
    $\begingroup$ Pythagorean triples and Hilbert’s Theorem 90 Noam D. Elkies I think this will be a nice short reference $\endgroup$ – Bumblebee Jul 7 '14 at 3:03
3
$\begingroup$

Actually, the complete rational solution to,

$$x^2+ny^2 = z^2$$

needs a third parameter, the scaling factor $t$, hence,

$$\big((a^2-nb^2)t\big)^2+n(abt)^2 = \big((a^2+nb^2)t\big)^2$$

Proof of completeness:

Let $xyz \ne 0$. Using a system of three equations,

$$\begin{aligned} x\,&=(a^2-nb^2)t\\ y\,&=abt\\ z\,&=(a^2+nb^2)t \end{aligned}$$

one can then rationally express the three unknowns $a,b,t$ in terms of the knowns $x,y,z$ by the simple formulas,

$$\begin{aligned} a\,&=x+z\\ b\,&=y\\ c\,&=\tfrac{1}{2(x+z)} \end{aligned}$$

though one needs a little algebraic manipulation to get these forms.

P.S. Just a minor quibble, $x^2-dy^2 =N$ is a Pell-like equation, while $ax^2+by^2+cz^2 = 0$ is a ternary quadratic form.

$\endgroup$
  • $\begingroup$ Yes, your scaling factor is Elkies’s "proportional to". $\endgroup$ – Kieren MacMillan Nov 29 '14 at 0:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.