I have a set of 29 coupled quadratic equations, with 29 unknown variables.
Can anyone offer any advice on how I could go about solving this?
3 days of staring at a wall has so far given me no thoughts on how to do this at all.
EDIT: \begin{align} T_1 &= X_1^2 X_2 X_3 X_4 X_5 X_6 \\ T_2 &= X_2^2 X_1 X_3 X_4 X_5 X_6 \\ T_3 &= X_3^2 X_1 X_2 X_4 X_5 X_6 \\ T_4 &= X_4^2 X_1 X_2 X_3 X_5 X_6 \\ T_5 &= X_5^2 X_1 X_2 X_3 X_4 X_6 \\ T_6 &= X_6^2 X_1 X_2 X_3 X_4 X_5 \\ T_7 &= X_1 X_2 X_3 X_4 X_5 X_6 X_7^2 X_8 X_9 X_{10} (1-X_5) \\ T_8 &= X_1 X_2 X_3 X_4 X_5 X_6 X_7 X_8^2 X_9 X_{10} (1-X_5) \\ T_9 &= X_1 X_2 X_3 X_4 X_5 X_6 X_7 X_8 X_9^2 X_{10} (1-X_5) \\ T_{10} &= X_1 X_2 X_3 X_4 X_5 X_6 X_7 X_8 X_9 X_{10}^2 (1-X_5) \\ T_{11} &= X_1 X_2 X_3 X_4 X_5 X_6 \{ (1-X_6) + (1-X_{10})(1-X_5)X_7 X_8 X_9 X_{10} \} X_{11}^2 X_{12} X_{13} X_{14} X_{15} \\ T_{12} &= X_1 X_2 X_3 X_4 X_5 X_6 \{ (1-X_6) + (1-X_{10})(1-X_5)X_7 X_8 X_9 X_{10} \} X_{11} X_{12}^2 X_{13} X_{14} X_{15} \\ T_{13} &= X_1 X_2 X_3 X_4 X_5 X_6 \{ (1-X_6) + (1-X_{10})(1-X_5)X_7 X_8 X_9 X_{10} \} X_{11} X_{12} X_{13}^2 X_{14} X_{15} \\ T_{14} &= X_1 X_2 X_3 X_4 X_5 X_6 \{ (1-X_6) + (1-X_{10})(1-X_5)X_7 X_8 X_9 X_{10} \} X_{11} X_{12} X_{13} X_{14}^2 X_{15} \\ T_{15} &= X_1 X_2 X_3 X_4 X_5 X_6 \{ (1-X_6) + (1-X_{10})(1-X_5)X_7 X_8 X_9 X_{10} \} X_{11} X_{12} X_{13} X_{14} X_{15}^2 \\ T_{16} &= X_1 X_2 X_3 X_4 X_5 X_6 (1-X_6)(1-X_9)X_7 X_8 X_9 X_{10} \} X_{11} X_{12} X_{13} X_{14} X_{15} X_{16}^2 X_{17} \\ T_{17} &= X_1 X_2 X_3 X_4 X_5 X_6 (1-X_6)(1-X_9)X_7 X_8 X_9 X_{10} \} X_{11} X_{12} X_{13} X_{14} X_{15} X_{16} X_{17}^2 \\ T_{18} &= X_1 X_2 X_3 X_4 X_5 X_6 X_7 X_8 X_9 X_{10} \} X_{11} X_{12} X_{13} X_{14} X_{15} X_{16} X_{17} (1-X_9)(1-X_5)(1-X_{16}) X_{18}^2 X_{19} X_{20} X_{21} \\ T_{19} &= X_1 X_2 X_3 X_4 X_5 X_6 X_7 X_8 X_9 X_{10} \} X_{11} X_{12} X_{13} X_{14} X_{15} X_{16} X_{17} (1-X_9)(1-X_5)(1-X_{16}) X_{18} X_{19}^2 X_{20} X_{21} \\ T_{20} &= X_1 X_2 X_3 X_4 X_5 X_6 X_7 X_8 X_9 X_{10} \} X_{11} X_{12} X_{13} X_{14} X_{15} X_{16} X_{17} (1-X_9)(1-X_5)(1-X_{16}) X_{18} X_{19} X_{20}^2 X_{21} \\ T_{21} &= X_1 X_2 X_3 X_4 X_5 X_6 X_7 X_8 X_9 X_{10} \} X_{11} X_{12} X_{13} X_{14} X_{15} X_{16} X_{17} (1-X_9)(1-X_5)(1-X_{16}) X_{18} X_{19} X_{20} X_{21}^2 \\ T_{22} &= X_1 X_2 X_3 X_4 X_5 X_6 \{ (1-X_5) X_7 X_8 X_9 X_{10} [(1-X_{17} + (1-X_9)(1-X_7)X_{16}X_{17}] + (1-X_2)\} X_{22}^2 X_{23} \\ T_{23} &= X_1 X_2 X_3 X_4 X_5 X_6 \{ (1-X_5) X_7 X_8 X_9 X_{10} [(1-X_{17} + (1-X_9)(1-X_7)X_{16}X_{17}] + (1-X_2)\} X_{22} X_{23}^2 \\ T_{24} &= X_1 X_2 X_3 X_4 X_5 X_6 \{ (1-X_3) + (1-X_5)(1-X_8)X_7 X_8 X_9 X_{10} \} X_{24}^2 X_{25} \\ T_{25} &= X_1 X_2 X_3 X_4 X_5 X_6 \{ (1-X_3) + (1-X_5)(1-X_8)X_7 X_8 X_9 X_{10} \} X_{24} X_{25}^2 \\ T_{26} &= X_1 X_2 X_3 X_4 X_5 X_6 \{ (1-X_4) + \{ (1-X_6) + (1-X_{10})(1-X_5))X_7 X_8 X_9 X_{10} X_{11}X_{12} X_{13} X_{14} X_{15} \}(1-X_{12}) + X_7 X_8 X_9 X_{10} X_{11}X_{12} X_{13} X_{14} X_{15}X_{16}X_{17}X_{18}X_{19}X_{20}X_{21}(1-X_{20})(1-X_5)(1-X_9)(1-x_16) +(1-X_25)\{(1-X_3) + (1-X_5)X_7 X_8 X_9 X_{10} (1-X_8) \}X_{24}X_{25} \} X_{26} \\ T_{27} &= X_1 X_2 X_3 X_4 X_5 X_6 \{ (1-X_4) + \{ (1-X_6) + (1-X_{10})(1-X_5))X_7 X_8 X_9 X_{10} X_{11}X_{12} X_{13} X_{14} X_{15} \}(1-X_{12}) + X_7 X_8 X_9 X_{10} X_{11}X_{12} X_{13} X_{14} X_{15}X_{16}X_{17}X_{18}X_{19}X_{20}X_{21}(1-X_{20})(1-X_5)(1-X_9)(1-x_16) +(1-X_25)\{(1-X_3) + (1-X_5)X_7 X_8 X_9 X_{10} (1-X_8) \}X_{24}X_{25} \} X_{27} \\ T_{28} &= \{ X_1 X_2 X_3 X_4 X_5 X_6 \{ (1-X_4) + \{ (1-X_6) + (1-X_{10})(1-X_5))X_7 X_8 X_9 X_{10} X_{11}X_{12} X_{13} X_{14} X_{15} \}(1-X_{12}) + X_7 X_8 X_9 X_{10} X_{11}X_{12} X_{13} X_{14} X_{15}X_{16}X_{17}X_{18}X_{19}X_{20}X_{21}(1-X_{20})(1-X_5)(1-X_9)(1-x_16)+(1-X_25)\{(1-X_3) + (1-X_5)X_7 X_8 X_9 X_{10} (1-X_8) \}X_{24}X_{25} \} \}(1-X_{27}+ X_1 X_2 X_3 X_4 X_5 X_6 X_7 X_8 X_9 X_{10} \} X_{11} X_{12} X_{13} X_{14} X_{15} X_{16} X_{17} (1-X_9)(1-X_5)(1-X_{16}) X_{18} X_{19} X_{20} X_{21}(1-X_20) +(1-X_{13})X_{28}X_{29} X_1 X_2 X_3 X_4 X_5 X_6 \{ (1-X_6) + (1-X_{10})(1-X_5)X_7 X_8 X_9 X_{10} \} X_{11} X_{12} X_{13} X_{14} X_{15} \} X_{28} \\ T_{29} &= \{ X_1 X_2 X_3 X_4 X_5 X_6 \{ (1-X_4) + \{ (1-X_6) + (1-X_{10})(1-X_5))X_7 X_8 X_9 X_{10} X_{11}X_{12} X_{13} X_{14} X_{15} \}(1-X_{12}) + X_7 X_8 X_9 X_{10} X_{11}X_{12} X_{13} X_{14} X_{15}X_{16}X_{17}X_{18}X_{19}X_{20}X_{21}(1-X_{20})(1-X_5)(1-X_9)(1-x_16)+(1-X_25)\{(1-X_3) + (1-X_5)X_7 X_8 X_9 X_{10} (1-X_8) \}X_{24}X_{25} \} \}(1-X_{27}+ X_1 X_2 X_3 X_4 X_5 X_6 X_7 X_8 X_9 X_{10} \} X_{11} X_{12} X_{13} X_{14} X_{15} X_{16} X_{17} (1-X_9)(1-X_5)(1-X_{16}) X_{18} X_{19} X_{20} X_{21}(1-X_20) +(1-X_{13})X_{28}X_{29} X_1 X_2 X_3 X_4 X_5 X_6 \{ (1-X_6) + (1-X_{10})(1-X_5)X_7 X_8 X_9 X_{10} \} X_{11} X_{12} X_{13} X_{14} X_{15} \} X_{29} \end{align}
Here are the equations, my unknowns are the $X$ terms. The $T$ terms are known.
Both $T$ and $X$ terms are real and positive.
Thanks for the comments, still attempting to understand what a “Groebner basis“ is so far… EDIT END
In the following, I assume that the $(X_i)$ are real numbers.
Following the Pp.. comment about Grobner basis, we can solve the system constituted with the first $6$ equations when the $(T_i)_{i\leq 6}$ are generically chosen.
We obtain a result in the form $a{X_1}^7+b=0$ and, for every $i\leq 6$, $X_i=c_iX_1$. If the $(T_i)$ and $(X_i)$ are real, then we obtain a sole (and explicit) solution for $(X_i)_{i\leq 6}$.
EDIT 1. Since the $(X_i)_{i\leq 6}$ are known, solve the following blocks of equations (we obtain exactly the same type of equations as above)
i) $\{7,\cdots,10\}$, $a{X_7}^5+b=0$, one solution in $X_7,\cdots,X_{10}$.
ii) $\{11\cdots 15\}$, $aX_{11}^6+b=0$, $0$ or $2$ solutions in $X_{11},\cdots,X_{15}$..
iii) $\{16,17\}$, $aX_{16}^3+b=0$, one solution.
iv) $\{18,\cdots,21\}$, after $\{22,23\}$, after $\{24,25\}$. Equations $26,27$ have degree $1$ in $X_{26},X_{27}$. Each of these equations admits generically a unique solution.
v) Equations $28,29$ are in the form $aX_{28}X_{29}+bX_{28}=c,dX_{28}X_{29}+eX_{29}=f$, $0$ or $2$ solutions.
EDIT 2. Grobner basis theory is useless. Indeed the solution(s) of a system of $n$ equations in the form of our first $6$ equations is: ${X_1}^{n+1}=\dfrac{{T_1}^n}{T_2\cdots T_n}$ and, for every $i\leq n$, $X_i=\dfrac{T_i}{T_1}X_1$.
EDIT 3. (answer to Respawned Fluff). The set of solutions of our system is zero-dimensional over $\mathbb{C}$. Then it is easy to keep only the real solutions. Here the system is block-triangular ; moreover each block (using the solutions of the previous blocks) admits an "effective solution" in the following sense: there is $i$ and a polynomial $P$ of degree $d$ s.t. $P(x_i)=0$ and, for every $j\not= i$, there are polynomials $P_j$ of degre $<d$ s.t. $x_j=P_j(x_i)$. Finally, our system has generically $0$ or $4$ real solutions. This is a simple system and clearly it can be easily solved by the standard softwares under the condition that the chosen order for the unknown is essentially $X_1,\cdots,X_{29}$ ; otherwise, the calculation time is likely to be very long.
At least, 2 teams of researchers are working about this subject: the LIP 6 laboratory (J.C. Faugère) and a group around M. Moreno Maza. The first one studies the so-called semi-regular systems over $\mathbb{C}$ and the second one studies the so-called regular semi-algebraic systems over $\mathbb{R}$.