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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 $$(u,v,w) = \biggl(\frac{d_2v_2^2+d_1v_1^2}{2},v_1v_2,\frac{d_2v_2^2-d_1v_1^2}{2}\biggr).$$

Bart Michels
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Kieren MacMillan
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    Obviously $y$ from $(\star),$is always even, so the solution $x=2, y=1$ for $d=3, z=1$ cannot be produced. – gammatester Jun 26 '14 at 13:25
  • 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. – Thomas Andrews Jun 26 '14 at 13:54
  • This formula really makes all the decisions. The formula has a General view there. http://math.stackexchange.com/questions/738446/solutions-to-ax2-by2-cz2/738527#738527 Interestingly, the formula for the equation $x^2+y^2=dz^2$ so simply cannot be recorded. It has a bulky look – individ Jun 26 '14 at 14:03
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    Just for reference of others, here's a link to the book, which is public domain: http://www.gutenberg.org/files/20073/20073-pdf.pdf – Thomas Andrews Jun 26 '14 at 14:07
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    Pythagorean triples and Hilbert’s Theorem 90 Noam D. Elkies
    I think this will be a nice short reference     – Bumblebee Jul 07 '14 at 03:03
  • So how does Elkies' paper jive with @gammatester's counterexample? His $A^2-4B$ would be $0^2-4(-3)=12$ which is not a square, so the parameterization should work, right? – Kieren MacMillan Jul 07 '14 at 13:12
  • Ah! If $(m,n)=1$, then you get $$(x',y',z')=(m^2+3n^2,2mn,m^2-3n^2)=(4,2,-2)$$ which is, as claimed, proportional to the desired $(x,y,z)=(2,1,-1)$ [here with $z=-1$ instead of $+1$].

    Any solution $(m,n)=(3n,n)$ will give you a solution proportional to $(2,1,1)$ [with $z$ positive. For example, $(m,n)=(3,1)$ yields $$(x',y',z')=(m^2+3n^2,2mn,m^2-3n^2)=(12,6,6),$$ and dividing by $6$ gives you the desired $(x,y,z)=(2,1,1)$.

    Yay, Noam!! =)

     – Kieren MacMillan Jul 07 '14 at 14:30

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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.

Tito Piezas III
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