3
$\begingroup$

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,

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

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Next 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
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$

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.