Solve the following system of Diophantine equations $$ \begin{cases} x^2+z^2=u^2,\\ y^2+4 z^2=v^2,\\ (x+y)^2+z^2=w^2. \end{cases} $$

By computer I can find some solutions $(x,y,z,u,v,w)$ for example $(25,119,60,65,169,156).$

Question: Is there any good parametrisation of the set of positibe integer solutions for the system?

I can find a parametrisation for first two equation

$$ \begin{cases} x^2+z^2=u^2,\\ y^2+4 z^2=v^2. \end{cases} $$ It follows that $(2x)^2+v^2=y^2+u^2$ and we have the equal sums of two squares wich can be parametrised in such a way $$ \begin{cases} 2x=m p-n q,\\ v=n p+m q, \\ y=m p+n q,\\ u=n p-m q. \end{cases} $$

Is is possyble to find a parametrisation of the system $$ \begin{cases} x^2+z^2=u^2,\\ y^2+4 z^2=v^2,\\ (x+y)^2+z^2=w^2? \end{cases} $$

  • $\begingroup$ It may make it easier if you eliminate $z^2$ and get just two equations in $(x,y,u,v,w)$. $\endgroup$ – George Law Jan 1 '17 at 10:15
  • $\begingroup$ For any $m,n$ you have that one set of solutions is $(x,y,z,u,v,w)=(n^2-m^2,2(m^2-n^2),2mn,m^2+n^2,2(m^2+n^2),m^2+n^2)$ $\endgroup$ – kingW3 Jan 1 '17 at 18:19
  • $\begingroup$ @kingW3, no, since yours $x$ and $y$ have different signs but I asked about positive integer solutions $\endgroup$ – Leox Jan 1 '17 at 18:58
  • $\begingroup$ where did you get the problem? $\endgroup$ – Will Jagy Jan 1 '17 at 20:36
  • $\begingroup$ It comes from a geometry problem about existing of some right triangles $\endgroup$ – Leox Jan 1 '17 at 21:24

We have \begin{equation*} \frac{x}{z}=\frac{e^2-1}{2e} \hspace{2cm} \frac{y}{2z}=\frac{f^2-1}{2f} \hspace{2cm} \frac{x+y}{z}=\frac{g^2-1}{2g} \end{equation*} where $e,f,g,\in \mathbb{Q}$.

Thus, we have the quadratic \begin{equation*} e^2+\frac{2f^2g+f(1-g^2)-2g}{fg}e-1=0 \end{equation*} which must hold. Since we wish $e \in \mathbb{Q}$, the discriminant of the quadratic must be a rational square, so there exists $d \in \mathbb{Q}$ such that \begin{equation*} d^2=4g^2f^4+4gf^3(1-g^2)+f^2(g^4-6g^2+1)+4gf(g^2-1)+4g^2 \end{equation*}

This quartic, in $f$, has a clear rational point when $f=0$, so is birationally equivalent to the elliptic curve \begin{equation*} V^2=U^3+(g^4+18g^2+1)U^2+64g^4U \end{equation*} with \begin{equation*} f=\frac{(g^2-1)U+V}{4g(4g^2+U)} \end{equation*}

The elliptic curve has a point of order $2$ at $(0,0)$ and two points of order $4$ when $U=-8g^2$. Numerical experiments suggest the rank is at least $2$, with $U=8$ giving one generator and $U=4g(g-1)^2$ giving a second.

These lead to an infinite number of possible parametric solutions. One simple one comes from $U=4g(g-1)^2, V=-4g(g^4-1)$, where \begin{equation*} x=2(g^6-4g^5+7g^4-12g^3+7g^2-4g+1) \hspace{1cm} y=2g(3g^4-8g^3+14g^2-8g+3) \hspace{1cm} z=4g(g+1)(g-1)(g^2-g+1) \end{equation*}

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  • $\begingroup$ Thank you but something is wrong, for example $$ y^2+4 z^2 =4\,{g}^{2} \left( {g}^{2}+1 \right) ^{2} \left( 5\,{g}^{2}-8\,g+5 \right) ^{2} \neq v^2=16 g^2(g^4-1)^2. $$ $\endgroup$ – Leox Jan 1 '17 at 17:19
  • $\begingroup$ Changed variable names to prevent confusion $\endgroup$ – Allan MacLeod Jan 1 '17 at 17:52
  • $\begingroup$ The one variable parametrisaion doesn't exhaust all solutions. Is it possyble to get a multivariate porametrisation or a set of one variable parametrisations to cover the whole set of solutions? $\endgroup$ – Leox Jan 1 '17 at 18:41

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