Find all rational solutions $(p,q,r)$ to the Diophantine equation $$\frac{2p}{1+2p-p^2}+\frac{2q}{1+2q-q^2}+\frac{2r}{1+2r-r^2}=1\,.$$ At least, determine an infinite family of $(p,q,r)\in\mathbb{Q}^3$ satisfying this equation. If you can also find such an infinite family of $(p,q,r)$ with the additional condition that $0<p,q,r<1+\sqrt{2}$, then it would be of great interest.
This question is related to Three pythagorean triples.
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If three rationals $a,b,c$ can be found, these lead to rational $p,q,r$. For simplicity, we'll start with the $a,b,c$. Given the system,
$$a^2+(b+c)^2 = x_1^2\\b^2+(a+c)^2 = x_2^2\\c^2+(a+b)^2 = x_3^2$$
Let $a = m^2-n^2,\;b=2mn-c,$ and it becomes,
$$(m^2+n^2)^2 = x_1^2\\2c^2+2c(m^2-2mn-n^2)+(m^2+n^2)^2 = x_2^2\\2c^2-2c(m^2+2mn-n^2)+(m^2+2mn-n^2)^2 = x_3^2$$
To simultaneously make two quadratics of form,
$$a_1x^2+a_2x+\color{blue}{a_3^2} = z_1^2\tag1$$
$$b_1x^2+b_2x+\color{blue}{b_3^2} = z_2^2\tag2$$
into squares is easy since their constant term is already a square. Assume,
$$a_1x^2+a_2x+\color{blue}{a_3^2} = (px+\color{blue}{a_3})^2$$
Expand and solve for $x$, and one finds,
$$x = \frac{a_2-2a_3p}{-a_1+p^2}\tag3$$
Substitute this into $(2)$ and one gets a quartic in $p$ with a square leading term. Since it is to be made a square, assume it to have the form,
$$\color{blue}{b_3^2}p^4+c_1p^3+c_2p^2+c_3p+c_4 = (\color{blue}{b_3}p^2+ep+f)^2$$
Expand and collect powers of $p$ to get,
$$d_1p^3+d_2p^2+d_3p+d_4 = 0$$
where the $d_i$ are polynomials in the two unknowns $e,f$. These two unknowns enable $d_1 = d_2 = 0$ to be solved, leaving only the linear eqn,
$$d_3p+d_4 = 0$$
Solve for $p$, substitute into $(3)$, and one gets an $x$ that simultaneously solves $(1)$ and $(2)$. Since one has $a,b,c$, then $p,q,r$ can be recovered using the relations described by the OP here.
For those not yet familiar with elliptic curves, this tangent method is easy to understand. It was known way back to Fermat, but has now been incorporated into the modern theory of elliptic curves. An illustrative example can also be found in this post.
If you combine the system into a single equation, you find that $r$ satisfies the quadratic equation \begin{equation*} r^2-\frac{4(p^2q+p(q^2-4q-1)-q)}{p^2(q^2-1)-4pq-q^2+1}r-1=0 \end{equation*}
For $r \in \mathbb{Q}$, the discriminant must be a rational square, so there must exist $d \in \mathbb{Q}$ such that \begin{equation*} d^2=(p^2+1)^2q^4-32p^2q^3+(2p^4-32p^3+68p^2+32p+2)q^2+32p^2q+(p^2+1)^2 \end{equation*}
The quartic has a rational solution when $q=0$, and so is birationally equivalent to an elliptic curve.
Using Mordell's method with a symbolic algebra package, we find this elliptic curve to be \begin{equation*} V^2=U^3+2(p^4-4p^3+10p^2+4p+1)U^2+(p^2-2p-1)^4U \end{equation*} with the reverse transformation \begin{equation*} q=\frac{(p^2+1)V+8p^2U}{(p^2+1)^2U+(p^2-2p-1)^4} \end{equation*}
Numerical experiments suggest that the torsion subgroup of the curve is isomorphic to $\mathbb{Z4}$, and we find $(0,0)$ is a point of order $2$, and $(-(p^2-2p-1)^2, \pm 4p(p^2-2p-1)^2)$ are of order $4$.
These experiments also suggest that the rank of the curves is always at least $1$, and it is not hard to find that $(-(p^2+1)^2, \pm 8p^2(p^2+1)\,)$ are of infinite order.
Taking the negative $V$ value gives \begin{equation*} q=\frac{2p(p^2+1)^2}{(p+1)(p-1)(p^4-2p^3+2p^2+2p+1)} \end{equation*} and, if we slot this into the quadratic for $r$ we get \begin{equation*} r=\frac{(p-1)(p^6-2p^5+7p^4-p^2+2p+1)}{(p+1)(p^6-2p^5-p^4+7p^2+2p+1)} \end{equation*}
Because there is a point of infinite order, we can generate an infinite number of parametric solutions, though they become increasingly complicated.
Hope this helps, and a Merry Christmas to all.