Equation $wxyz(w+x+y+z)=1$ in $\mathbb{Q}_+^4$

Question

In this thread on Math.SE, Noam D. Elkies give the following parametric family of solutions in $\mathbb{Q_+}^3$ of the equation $xyz(x+y+z)=1$ (found by Euler) :

$$ x = \frac{6 t^3 (t^4-2)^2} {(4 t^4 + 1) (2 t^8 + 10 t^4 - 1)}, $$ $$ y = \frac{ 3 (4 t^4 + 1)^2} {2t (t^4-2) (2 t^8 + 10 t^4 - 1)}, $$ $$ z = \frac{ 2 (2 t^8 + 10 t^4 - 1)} {3t (4 t^4 + 1)}. $$

Can we do the same for the equation $wxyz(w+x+y+z)=1$ ?

Thank you.

Remarks :

1) I crossposted to Math.SE.

2) A computer investigation leads to the following solutions $(w\,;x\,;y\,;z)$ :

$\left(\dfrac{1}{6}\,;\dfrac{1}{3}\,;2\,;2\right)$ , $\left(\dfrac{1}{6}\,;\dfrac{2}{3}\,;\dfrac{2}{3}\,;3\right)$ , $\left(\dfrac{1}{12}\,;\dfrac{1}{4}\,;\dfrac{2}{3}\,;8\right)$ , $\left(\dfrac{1}{12}\,;\dfrac{1}{4}\,;\dfrac{8}{3}\,;3\right)$ , $\left(\dfrac{1}{12}\,;\dfrac{1}{3}\,;1\,;\dfrac{16}{3}\right)$ , $\left(\dfrac{1}{10}\,;\dfrac{2}{5}\,;2\,;\dfrac{5}{2}\right)$ , $\left(\dfrac{1}{14}\,;1\,;\dfrac{7}{4}\,;\dfrac{7}{4}\right)$ , $\left(\dfrac{1}{30}\,;\dfrac{5}{6}\,;\dfrac{9}{5}\,;\dfrac{10}{3}\right)$ , $\left(\dfrac{1}{24}\,;\dfrac{9}{8}\,;\dfrac{3}{2}\,;\dfrac{8}{3}\right)$ , $\left(\dfrac{1}{4}\,;\dfrac{1}{4}\,;\dfrac{4}{3}\,;\dfrac{8}{3}\right)$ , $\left(\dfrac{1}{10}\,;\dfrac{2}{3}\,;\dfrac{9}{10}\,;\dfrac{10}{3}\right)$ , $\left(\dfrac{1}{10}\,;\dfrac{5}{6}\,;\dfrac{16}{15}\,;\dfrac{5}{2}\right)$ , $\left(\dfrac{1}{10}\,;\dfrac{16}{15}\,;\dfrac{3}{2}\,;\dfrac{3}{2}\right)$ , $\left(\dfrac{1}{10}\,;\dfrac{6}{5}\,;\dfrac{6}{5}\,;\dfrac{5}{3}\right)$ , $\left(\dfrac{1}{6}\,;\dfrac{3}{10}\,;\dfrac{6}{5}\,;\dfrac{10}{3}\right)$ , $\left(\dfrac{1}{6}\,;\dfrac{3}{5}\,;1\,;\dfrac{12}{5}\right)$ , $\left(\dfrac{2}{9}\,;\dfrac{1}{4}\,;\dfrac{16}{9}\,;\dfrac{9}{4}\right)$ , $\left(\dfrac{3}{10}\,;\dfrac{5}{6}\,;1\,;\dfrac{6}{5}\right)$ , $\left(\dfrac{2}{5}\,;\dfrac{3}{5}\,;\dfrac{5}{6}\,;\dfrac{3}{2}\right)$ , $\left(\dfrac{1}{3}\,;\dfrac{5}{12}\,;\dfrac{5}{4}\,;\dfrac{8}{5}\right)$ , $\left(\dfrac{5}{12}\,;\dfrac{1}{2}\,;\dfrac{16}{15}\,;\dfrac{27}{20}\right)$ and $\left(\dfrac{9}{20}\,;\dfrac{2}{3}\,;\dfrac{5}{6}\,;\dfrac{5}{4}\right)$.

3) I found the following parametrization but $w$, $x$, $y$, and $z$ are never simultaneously positive :

$$w=\frac{296352\,t^4}{(2401\,t^5-16)(2401t^5-4)(2401\,t^5+8)}$$ $$x=-\frac{(2401\,t^5-4)^2}{21\,t\,(2401\,t^5-16)}$$ $$y=\frac{(2401\,t^5-16)^2}{168\,t\,(2401\,t^5+8)}$$ $$z=\frac{2(2401\,t^5+8)}{7\,t\,(2401\,t^5-4)}$$

Answer 1

Let
\begin{align*} w &= \frac{c_1t^{t_1}}{ABC}\\ x &= \frac{c_2B^2t^{t_2}}{A}\\ y &= \frac{c_3A^2t^{t_3}}{C}\\ z &= \frac{c_4Ct^{t_4}}{B} \end{align*} Then $$wxyz(w+x+y+z) = \frac{c_1t^{t_1}c_2t^{t_2}c_3t^{t_3}c_4t^{t_4}(c_1t^{t_1}+c_2B^3t^{t_2}C+c_3A^3t^{t_3}B+c_4C^2t^{t_4}A)}{C^2AB}$$

Hence
\begin{align*} &(c_1t^{t_1}c_2t^{t_2}c_3t^{t_3}c_4^2(t^{t_4})^2A-AB)C^2\\ &+c_1t^{t_1}c_2^2(t^{t_2})^2c_3t^{t_3}c_4t^{t_4}B^3C\\ &+c_1t^{t_1}c_2t^{t_2}c_3t^{t_3}c_4t^{t_4}(c_1t^{t_1}+c_3A^3t^{t_3}B)=0 \end{align*}

Since $C$ must be rational number then discriminant must be square number.
\begin{align*} v^2 &= (4c_3t^{t_3}B^2-4Bc_3^2(t^{t_3})^2c_1t^{t_1}c_2t^{t_2}c_4^2(t^{t_4})^2)c_1t^{t_1}c_2t^{t_2}c_3t^{t_3}c_4t^{t_4}A^4\\ &+(4Bc_1t^{t_1}-4c_1^2(t^{t_1})^2c_2t^{t_2}c_3t^{t_3}c_4^2(t^{t_4})^2)c_1t^{t_1}c_2t^{t_2}c_3t^{t_3}c_4t^{t_4}A\\ &+c_1^2(t^{t_1})^2c_2^4(t^{t_2})^4c_3^2(t^{t_3})^2c_4^2(t^{t_4})^2B^6 \end{align*}

Obviously, quartic is birationally equivalent to an elliptic curve.

For instance, take $(c_1,c_2,c_3)=(1,1,1),(t1,t2,t3)=(1,1,1),B=1$, then

$$v^2 = (4t^5-4t^{10})A^4+(4t^5-4t^{10})A+t^{10}$$

Quartic is transformed to an elliptic curve below.

$$E: Y^2+(4-4t^5)YX = X^3+(-4+8t^5-4t^{10})X^2+(-16t^{15}+16t^{20})X+64t^{15}-192t^{20}+192t^{25}-64t^{30}$$

$E$ has a point $P(X,Y)=(4-8t^5+4t^{10}, -16+48t^5-48t^{10}+16t^{15})$.
We obtain $$2P(X)=\frac{1-26t^{15}+33t^{20}-24t^{25}+8t^{30}-6t^5+15t^{10}}{(t^{10}-2t^5+1)^2}$$
According to Nagell-Lutz theorem, the point $P(X,Y)$ is not a point of finite order, hence we can obtain infinitely many parametric solutions.

Thus we can obtain a quartic point $$2Q(A) = \frac{-2t^5(t-1)^2(t^4+t^3+t^2+t+1)^2}{(2t^5-1)(t^2-t+1)(t^8+t^7-t^5-t^4-t^3+t+1))}$$

Then we obtain $(A,C)$

$$A = \frac{-2t^5(t-1)^2(t^4+t^3+t^2+t+1)^2}{(2t^5-1)(t^2-t+1)(t^8+t^7-t^5-t^4-t^3+t+1)}$$ $$C = \frac{-(3t^{10}-3t^5+1)}{-1+3t^5-3t^{10}+2t^{15}}$$

Finally we obtain $(w,x,y,z)$ \begin{align*} w &= \frac{1}{2}\frac{(2t^5-1)^2(t^2-t+1)^2(t^8+t^7-t^5-t^4-t^3+t+1)^2}{t^4(t-1)^2(t^4+t^3+t^2+t+1)^2(3t^{10}-3t^5+1)}\\ x &= \frac{-1}{2}\frac{(2t^5-1)(t^2-t+1)(t^8+t^7-t^5-t^4-t^3+t+1)}{t^4(t-1)^2(t^4+t^3+t^2+t+1)^2}\\ y &= \frac{-4(t^4+t^3+t^2+t+1)^4(t-1)^4t^{11}}{(3t^{10}-3t^5+1)(2t^5-1)(t^2-t+1)(t^8+t^7-t^5-t^4-t^3+t+1)}\\ z &= \frac{-(3t^{10}-3t^5+1)t}{(2t^5-1)(t^2-t+1)(t^8+t^7-t^5-t^4-t^3+t+1)} \end{align*}

This parametrization gives positive solution where $0\lt t \lt 1/2^{1/5}$.

Needless to say, this method can be applied to $xyz(x+y+z)=a$.

Answer 2

I finally found the following parametrization :

$$w=\frac{32916734851\,t^4}{(219896613895728\,t^5+1)(370352191824384\,t^5+1)(479774430317952\,t^5+1)}$$ $$x=\frac{32\,(219896613895728\,t^5+1)(479774430317952\,t^5+1)}{5187\,t\,(370352191824384\,t^5+1)}$$ $$y=\frac{3\,(370352191824384\,t^5+1)}{1232\,t\,(479774430317952\,t^5+1)}$$ $$z=\frac{(370352191824384\,t^5+1)(479774430317952\,t^5+1)}{4368\,t\,(219896613895728\,t^5+1)}$$

In computer-readable format :

w:=32916734851*t^4/((219896613895728*t^5+1)*(370352191824384*t^5+1)*(479774430317952*t^5+1))

x:=32*(219896613895728*t^5+1)*(479774430317952*t^5+1)/(5187*t*(370352191824384*t^5+1))

y:=3*(370352191824384*t^5+1)/(1232*t*(479774430317952*t^5+1))

z:=(370352191824384*t^5+1)*(479774430317952*t^5+1)/(4368*t*(219896613895728*t^5+1))

Answer 3

My idea is to try to find a polynomial equality of the form $$X+\lambda P^2Q^2+\mu PR^2+\varepsilon Q^2R^2=\theta P^2QR \;\;\;(1)$$ with $\lambda$, $\mu$, $\varepsilon$, $\theta$ in $\mathbb{Q}$ and $P$, $Q$, $R$ in $\mathbb{Q}[X] $.

Let $A=X$, $B=\lambda P^2Q^2$, $C=\mu PR^2$ and $D=\varepsilon Q^2R^2$, then $$ABCD(A+B+C+D)=\lambda\mu\varepsilon\theta X(PQR)^5 \;\;\;(E_a)\;.$$ Taking $X=kat^5$ with $k\in\mathbb{Q}$ such that $k\lambda\mu\varepsilon\theta=K^5$ with $K\in\mathbb{Q }$, we have $$wxyz(w+x+y+z)=a \;\;\;(E_a)\;,$$ with $w=\dfrac{A}{KtPQR}$, $x=\dfrac{B}{KtPQR}$, $w=\dfrac{C}{KtPQR}$ and $w=\dfrac{D}{KtPQR }$.

In fact, we will see that we can find an equality of the form $(1)$ with $P$, $Q$ and $R$ of degree 1.

Take $P=X-\alpha$, $Q=X-\beta$ and $R=X-\gamma$ with $\alpha$, $\beta$ and $\gamma$ in $\mathbb{Q}$ (such that $\gamma\neq\beta$).

Suppose $(1)$ holds ; by replacing $X$ by $\alpha$, $\beta$ then $\gamma$, by derivating then by replacing $X$ by $\alpha$, and by considering the dominant terms, we obtain

$$\left\{\begin{array}{l}\alpha+\varepsilon(\alpha-\beta)^2(\alpha-\gamma)^2=0\\\beta+\mu(\beta-\alpha)(\beta-\gamma)^2=0\\\gamma+\lambda(\gamma-\alpha)^2(\gamma-\beta)^2=0\\1+\mu(\alpha-\gamma)^2+2\varepsilon(\alpha-\beta)(\alpha-\gamma)(2\alpha-\gamma-\beta)=0\\\theta=\lambda+\varepsilon\\\end{array}\right.$$ which is equivalent to $$\left\{\begin{array}{l} \varepsilon=-\dfrac{\alpha}{(\alpha-\beta)^2(\alpha-\gamma)^2}\\ \mu=-\dfrac{\beta}{(\beta-\alpha)(\beta-\gamma)^2}\\ \lambda=-\dfrac{\gamma}{(\gamma-\alpha)^2(\gamma-\beta)^2}\\ (\alpha-\beta)(\alpha-\gamma)(\beta-\gamma)^2+\beta(\alpha-\gamma)^3-2\alpha(\beta-\gamma)^2(2\alpha-\gamma-\beta) \;\;\;(2)\\ \theta=\lambda+\varepsilon\\ \end{array}\right.$$ Let $u=\dfrac{\alpha-\gamma}{\gamma-\beta}$; the identity $(2)$ gives $$(\alpha-\beta)u(\gamma-\beta)^3+\beta u^3(\gamma-\beta)^3-2\alpha(\gamma-\beta)^2(2( \alpha-\gamma)+\gamma-\beta) \;,$$ which is equivalent to $$(\alpha-\beta)u+\beta u^3-2\alpha(2u+1)=0 \;,$$ which leads to $$\left\{\begin{array}{l} \beta=\dfrac{3u+2}{u^3-u}\alpha\\ \gamma=\dfrac{u^2+3u+1}{(1+u)(u^2-1)}\alpha\\ \varepsilon=-\dfrac{u^2(u+1)^6(u-1)^4}{\alpha^3(u^3-4u-2)^4}\\ \mu=\dfrac{u^2(u+1)^4(u-1)^2(3u+2)}{\alpha^2(u^3-4u-2)^3}\\ \lambda=-\dfrac{u^2(u+1)^6(u-1)^3(u^2+3u+1)}{\alpha^3(u^3-4u-2)^4}\\ \theta=-\dfrac{u^3(u+1)^6(u-1)^3(u+4)}{\alpha^3(u^3-4u-2)^4}\\ \end{array}\right.$$ Conversely, one can verify that if $\beta$, $\gamma$, $\varepsilon$, $\mu$, $\lambda$ and $\theta$ are defined in terms of $u$ and $\alpha$ by the previous relations, then the polynomial relation $(1)$ is satisfied.

Taking $k=-\dfrac{\alpha u}{(u+1)^2(u-1)^2(u^2+3u+1)(u+4)(3u+2)}$, we have $k\lambda\mu\varepsilon\theta=K^5$ with $K=\dfrac{u^2(u+1)^4(u-1)^2}{\alpha^2(u^ 4-4u-2)^3}$; we then choose $X=kat^5$. Let's set $$\left\{\begin{array}{l} F(u)=(u+1)^2(u-1)^2(u^2+3u+1)(u+4)(3u+2)\\ G(u)=(u+1)(u-1)(u^2+3u+1)(u+4)(3u+2)^2\\ H(u)=(u-1)(u^2+3u+1)^2(u+4)(3u+2)\\ J(u)=(u+4)(3u+2)(u^3-4u-2)\\ \end{array}\right.$$ we obtain (after simplifications) the following $2-$parameterization of an infinite subset of solutions of $(E_a)$: $$\left\{\begin{array}{l} w=\dfrac{a(u-1)^2(u^2+3u+1)^2(u^3-4u-2)J(u)^2t^4}{(aut^5+F(u))(au^2t^5+G(u))(aut^5+H(u))}\\ x=\dfrac{(aut^5+F(u))(au^2t^5+G(u))}{u(u-1)J(u)t(aut^5+H(u))}\\ y=\dfrac{u(3u+2)(aut^5+H(u))}{t(au^2t^5+G(u))}\\ z=\dfrac{(au^2t^5+G(u))(aut^5+H(u))}{u(u^2+3u+1)J(u)t(aut^5+F(u))}\\ \end{array}\right.$$ By choosing for example $a=1$ and $u=3$, we obtain the following $1-$parametrization of an infinite subset of solutions of $(E_1)$ $$\left\{\begin{array}{l} w=\dfrac{18809562772t^4}{(3t^5+55594)(3t^5+93632)(9t^5+128744)}\\ x=\dfrac{(3t^5+93632)(9t^5+128744)}{6006t(3t^5+55594)}\\ y=\dfrac{33(3t^5+55594)}{t(9t^5+128744)}\\ z=\dfrac{(3t^5+55594)(9t^5+128744)}{57057t(3t^5+93632)}\\ \end{array}\right.$$