The problem is to solve the equation,

$$\frac{(1-x)^2}{x} + \frac{(1-y)^2}{y} + \frac{(1-z)^2}{z} + 4 = 0\tag{1}$$

in the rationals. Treating this as an equation in $z$, easy solutions would involve $z = \pm 1$, $z = \pm x, \pm y$. More complicated ones would be to make the discriminant $D$ of $(1)$ a square,

$$D = -4x^2y^2 + (x+y-2xy+x^2y+xy^2)^2 = t^2\tag{2}$$

with one solution being,

$$ x = a/b$$

$$ y = -b/a\, (p_1/p_2)$$

$$p_1 = 7a^6 + 4a^5b - 14a^4b^2 + 12a^3b^3 - 25a^2b^4 + 8a b^5 - 8b^6$$

$$p_2 = 8a^6 - 8a^5b + 25a^4b^2 - 12a^3b^3 + 14a^2b^4 - 4a b^5 - 7b^6$$

and $a,b$ being the legs of the Pythagorean triple $a^2+b^2 = c^2$.

However, can someone find a polynomial parameterization of small degree to $(1)$?

Edit (a few days later):

Courtesy of Allan MacLeod, a simple parameterization to $(1)$ can be given by using $x=1/y$ and the discriminant $(2)$ greatly simplifies to just solving,

$$y^2+1 = w^2$$

Hence, his first answer below can also be expressed in terms of Pythagorean triples $a^2+b^2 = c^2$ as,

$$x=a/b,\;\; y = b/a,\;\; z = \frac{(a+c)(b-c)}{ab}$$

  • $\begingroup$ If $a=3,b=4,$ then $p_1=-54209,p_2=58360,x=3/4,y=54209/43770,z=\frac{-3368103939\pm\sqrt{1335523557592498121}}{3163637240}\not\in\mathbb Q$. Am I wrong? $\endgroup$ – lsr314 Nov 17 '13 at 7:33
  • $\begingroup$ @Hecke: You must have mis-typed $p_2$. Using the polynomial above, I get instead $p_2 = 1016$, (and $y = 54209/762$). $\endgroup$ – Tito Piezas III Nov 17 '13 at 15:32

After some further thought and computation, if $f=(k^2-1)/2k$ then $u=2(k-1)^2$ gives

\begin{equation*} v= \pm \frac{(k-1)^2(k^2-2k-1)(k^2-2k+3)}{2k^2} \end{equation*}

This point gives the parametric form \begin{equation*} f=\frac{k^2-1}{2k} \hspace{2cm} g=\frac{2k}{k^2-1} \hspace{2cm} h=\frac{k(1-k)}{k+1} \end{equation*}

Doubling this point (with $f=(k^2-1)/2k$) gives \begin{equation*} g=\frac{(1-k)(k^2+2k-1)(3k^2+2k+1)}{2k(k+1)(k^2+1)(k^2-2k+3)} \end{equation*}

\begin{equation*} h=\frac{k(k+1)(k^2+2k-1)(k^2-2k+3)}{2(k-1)(k^2+1)(3k^2+2k+1)} \end{equation*}

Further forms can probably be derived using the torsion points.

  • $\begingroup$ Thanks. I should have checked the case $x = 1/y$. By the way, I assume you are the Allan MacLeod who also worked on $a^4+b^4+c^4 = d^4$? You may be interested in my MSE post and MO post on finding more Elkies-type elliptic curves for that equation. $\endgroup$ – Tito Piezas III Nov 20 '13 at 23:15

Consider the problem written as \begin{equation*} \frac{(1-f)^2}{f}+\frac{(1-g)^2}{g}+\frac{(1-h)^2}{h}+4=0 \end{equation*} which gives the quadratic in $h$ \begin{equation*} h^2+\frac{f^2g+f(g-1)^2+g}{f\,g}+1=0 \end{equation*}

As Tito stated, for this to have a rational solution (assuming $f,g$ rational) the discriminant must be a rational square. After simplification, this means there must be $t \in \mathbb{Q}$ with \begin{equation*} t^2=f^2g^4+2f(f^2-1)^2g^3+(f^4-4f^3+4f^2-4f+1)g^2+2f(f-1)^2g+f^2 \end{equation*}

Defining $Y=t\,f$ and $X=g\,f$ gives the quartic \begin{equation*} Y^2=X^4+2(f-1)^2X^3+(f^4-4f^3+4f^2-4f+1)X^2+2(f^2-f)^2X+f^4 \end{equation*}

There is an obvious rational point $X=0, Y = f^2$, so the curve is birationally equivalent to an elliptic curve. Using Mordell's method the elliptic curve is \begin{equation*} v^2=u^3+(f^4-4f^3-2f^2-4f+1)u^2+16f^4u \end{equation*} with \begin{equation*} g=\frac{u(f-1)^2-v}{2f(4f^2-u)} \end{equation*}

The curve has a point of order $2$ at $u=0$, two points order $4$ when $u=4f^2$ and $4$ points of order $8$ at $u=4f$ and $u=4f^3$. These give undefined solutions or permutations of $(f,-f,1)$ or $(f,-1/f,1)$. Thus the torsion subgroup is usually isomorphic to $\mathbb{Z}8$. There are $3$ rational points of order $2$ when $f^2-6f+1=\square$ when the torsion subgroup is $\mathbb{Z}2 \times \mathbb{Z}8$.

Numerical tests show that the elliptic curve often has rank $0$, and so no other solutions. For $f=10$, the rank is $1$ with generator $(-125,-8250)$ which gives $g=-5/28$ and $h= -7/4,-4/7$.

If we set $f=(k^2-1)/2k$, the parametric solution quoted comes from \begin{equation*} u=\frac{(k-1)^2(k^6+2k^5-13k^4-4k^3-5k^2-6k-7)^2}{2(2k^6-4k^5+5k^4-12k^3+4k^2+1)^2} \end{equation*}

Numerical tests suggest that this is often $3$ times a generator point. Thus, it might be possible to find a smaller parametric solution by doing some further algebra.


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