Okay, so I was trying to solve this riddle found here.
It is a diagram of a star with 16 points. Each point corresponds uniquely to a number between 1 and 16. The letters on each point represent a letter of some saying, where if the unique corresponding to a point is $n$, then the letter on that point is the $n$th letter of the saying. We are also given the condition that the sum of the 4 numbers on a given line segment is the same.
I noticed that $\sum^{16}_{i=1} i = 136$, and each point is counted exactly twice, so each line should add up to $2\cdot136/8 = 34$. So we can find 8 equations in this way.
$$x_1 + x_2 + x_3 + x_4 = 34 $$ $$x_4 + x_5 + x_6 + x_7 = 34 $$ $$x_2 + x_7 + x_8 + x_9 = 34 $$ $$x_5 + x_9 + x_{10} + x_{11} = 34 $$ $$x_8 + x_{11} + x_{12} + x_{13} = 34 $$ $$x_{10} + x_{13} + x_{14} + x_{15} = 34 $$ $$x_3 + x_{12} + x_{15} + x_{16} = 34 $$ $$x_1 + x_6 + x_{14} + x_{16} = 34 $$
So, I have 8 equations and 16 unknowns, and finding the coefficient matrix and putting it in rref didn't shed that much light on the matter, because there are still too many unknowns.
Now, I know there will be more than one solution. We can rotate it 7 times and there are 8 lines of symmetry; however, I think we should be able to narrow down our solutions farther than what I have already done.
Does anyone have other ideas on how to approach this problem?
Okay, the solution to this riddle is actually written in the blog you found it, but since you are asking about ideas on how to approach, i'll give u mine.
I found this solution (using GLPK) $$ \begin{array}{cccc} x_1 = 13 & x_2 = 8 & x_3 = 7 & x_4 = 6 \\ x_5 = 12 & x_6 = 2 & x_7 = 14 & x_8 = 11 \\ x_9 = 1 & x_{10} = 5 & x_{11} = 16 & x_{12} = 3 \\ x_{13} = 4 & x_{14} = 10 & x_{15} = 15 & x_{16} = 9 \end{array} $$ where im using the same notation you did above. I also tried to figure out what was that saying, but i get a nosense group of letters, which I guess its because there exist, as u noticed, at least 8 simetric solutions. I saw in the blog some people solved the riddle just looking at the letters, but I am more of numbers, so I'll just skip that "saying" part.
Linear algebra part
My aproach consists on using the 8 conditions you already wrote, but adding the condition that each number between 1 and 16 must appear exactly once. Also we have to face the problem that our variables are integers, not real numbers. I dont even know if its possible to get such conditions using only linear algebra equations, a start would be adding this $$ \sum_{i=1}^{16} x_i = 136 $$ but we still lack equations and i dont know how to proceed. So I have used linear programming, and GLPK to solve it given this conditions.
Linear programming and aproach used (a bit long)
I dont know if you are familiar with linear programing, but I will explain the procedure in a way that it can be understood, hopefully.
Basicaly we define a new set of variables $x_{ij}$ which are binary (i.e $0$ or $1$) and are related to your variables this way $$ x_i = \sum_{j=1}^{16} x_{ij}\cdot j $$ Of course, then we only allow one of this $x_{ij}$ to be diferent from zero for each $i$, becasue "every unknown is only one number" so we add the condition $$ \sum_{j=1}^{16} x_{ij} = 1 $$ The same way, we also want "each number to appear only once", which gives a very similar condition (note the change on the index of the sum) $$ \sum_{i=1}^{16} x_{ij} = 1 $$ Finally, we substitute this new $x_{ij}$ on the 8 equations we had at the start and run GLPK. It tries to find a feasible solution, and it succeeds, giving the one i wrote above. I guess on diferent machines the program might find diferent solutions, since we already know its not unique.
Extra (I almost found the saying)
I was a bit dissapointed i didnt get the saying, so i decided to cheat a bit, took a look on the blog, the text is "do a good turn daily". His solution involves Y being the last letter, which means $x_{12}=16$. So i put this restraint on my program and... still got a diferent solution. $$ \begin{array}{cccc} x_1 = 8 & x_2 = 4 & x_3 = 15 & x_4 = 7 \\ x_5 = 6 & x_6 = 12 & x_7 = 9 & x_8 = 10 \\ x_9 = 11 & x_{10} = 14 & x_{11} = 3 & x_{12} = 16 \\ x_{13} = 5 & x_{14} = 13 & x_{15} = 2 & x_{16} = 1 \end{array} $$ Not sure how many solutions this annoying star has :)