I'm hoping to find an analytical solution to this simple system of quadratic equation: $$ r_{1}^{2}=\theta_{1}^{2}\left(r_{2}+r_{3}\right)\\ r_{2}^{2}=\theta_{2}^{2}\left(r_{1}+r_{3}\right)\\ r_{3}^{2}=\theta_{3}^{2}\left(r_{1}+r_{2}\right) $$ where the $\theta_j$ are known parameters and we must solve for $r_j$. I've tried guessing some functional forms but with no luck. If we restrict the system to be two-dimensional ($\theta_3=0$) the solution can easily be computed to be $$ r_{1} =\left(\theta_{1}\theta_{2}^{0.5}\right)^{\frac{1}{1-0.5^{2}}}\\ r_{2} =\left(\theta_{2}\theta_{1}^{0.5}\right)^{\frac{1}{1-0.5^{2}}} $$ I haven't been able to find a way to generalize these solutions.
Anybody has any suggestions?
Edit: I should add that a solution could be found by iterating from an initial vector $r$.
Take any $r_3$ that satisfies the septic $$ {r_{{3}}}^{7}+ \left( -2\,{\theta_{{1}}}^{2}{\theta_{{3}}}^{4}-2\,{ \theta_{{2}}}^{2}{\theta_{{3}}}^{4} \right) {r_{{3}}}^{4}-3\,{\theta_{ {1}}}^{2}{\theta_{{2}}}^{2}{\theta_{{3}}}^{4}{r_{{3}}}^{3}-8\,{\theta_ {{1}}}^{2}{\theta_{{2}}}^{2}{\theta_{{3}}}^{6}{r_{{3}}}^{2}+ \left( -{ \theta_{{1}}}^{4}{\theta_{{2}}}^{2}{\theta_{{3}}}^{6}+{\theta_{{1}}}^{ 4}{\theta_{{3}}}^{8}-{\theta_{{1}}}^{2}{\theta_{{2}}}^{4}{\theta_{{3}} }^{6}-2\,{\theta_{{1}}}^{2}{\theta_{{2}}}^{2}{\theta_{{3}}}^{8}+{ \theta_{{2}}}^{4}{\theta_{{3}}}^{8} \right) r_{{3}}-2\,{\theta_{{1}}}^ {4}{\theta_{{2}}}^{2}{\theta_{{3}}}^{8}-2\,{\theta_{{1}}}^{2}{\theta_{ {2}}}^{4}{\theta_{{3}}}^{8} =0$$ Then $r_1$ can be found in terms of $r_3$: $$ \eqalign{r_1 &= \frac{a_1 r_3 + a_2 r_3^2 + a_3 r_3^3 + a_4 r_3^4 + a_5 r_3^5 + a_6 r_3^6 + a_7 r_3^7}{D}\cr a_1 &= {\theta_{{1}}}^{6}{\theta_{{3}}}^{6}-9\,{\theta_{{1}}}^{4}{\theta_{{2} }}^{2}{\theta_{{3}}}^{6}+8\,{\theta_{{1}}}^{4}{\theta_{{3}}}^{8}-9\,{ \theta_{{1}}}^{2}{\theta_{{2}}}^{4}{\theta_{{3}}}^{6}-16\,{\theta_{{1} }}^{2}{\theta_{{2}}}^{2}{\theta_{{3}}}^{8}+{\theta_{{2}}}^{6}{\theta_{ {3}}}^{6}+8\,{\theta_{{2}}}^{4}{\theta_{{3}}}^{8} \cr a_2 &= {\theta_{{1}}}^{6}{\theta_{{3}}}^{4}-4\,{\theta_{{1}}}^{4}{\theta_{{2} }}^{2}{\theta_{{3}}}^{4}+12\,{\theta_{{1}}}^{4}{\theta_{{3}}}^{6}-5\,{ \theta_{{1}}}^{2}{\theta_{{2}}}^{4}{\theta_{{3}}}^{4}-48\,{\theta_{{1} }}^{2}{\theta_{{2}}}^{2}{\theta_{{3}}}^{6}+4\,{\theta_{{2}}}^{4}{ \theta_{{3}}}^{6} \cr a_3 &= -2\,{\theta_{{1}}}^{4}{\theta_{{3}}}^{4}-28\,{\theta_{{1}}}^{2}{\theta _{{2}}}^{2}{\theta_{{3}}}^{4}-2\,{\theta_{{2}}}^{4}{\theta_{{3}}}^{4} \cr a_4 &= -{\theta_{{1}}}^{4}{\theta_{{3}}}^{2}-2\,{\theta_{{1}}}^{2}{\theta_{{2 }}}^{2}{\theta_{{3}}}^{2}-16\,{\theta_{{1}}}^{2}{\theta_{{3}}}^{4}-{ \theta_{{2}}}^{4}{\theta_{{3}}}^{2}-16\,{\theta_{{2}}}^{2}{\theta_{{3} }}^{4} \cr a_5 &= -4\,{\theta_{{1}}}^{2}{\theta_{{3}}}^{2}-4\,{\theta_{{2}}}^{2}{\theta_ {{3}}}^{2} \cr a_6 &= 2\,{\theta_{{1}}}^{2}+2\,{\theta_{{2}}}^{2}\cr a_7& = 8\cr D &= {\theta_{{1}}}^{6}{\theta_{{3}}}^{6}+{\theta_{{1}}}^{4}{\theta_{{2}}}^ {2}{\theta_{{3}}}^{6}+8\,{\theta_{{1}}}^{4}{\theta_{{3}}}^{8}-{\theta_ {{1}}}^{2}{\theta_{{2}}}^{4}{\theta_{{3}}}^{6}-{\theta_{{2}}}^{6}{ \theta_{{3}}}^{6}-8\,{\theta_{{2}}}^{4}{\theta_{{3}}}^{8} \cr} $$ (assuming of course $D \ne 0$), and finally $$ r_2 = \frac{r_3^2}{\theta_3^2} - r_1$$ I found this using a Groebner basis.