A system of equations is

$$x_1+x_2+x_3+x_4=b_1$$ $$x_1+x_5+x_6+x_7=b_2$$ $$x_2+x_5+x_7+x_8=b_3$$ $$x_3+x_6+x_8+x_{10}=b_4$$ $$x_4+x_7+x_9+x_{10}=b_5$$ $b_1,...,b_5$ and $x_1,...,x_{10}$ are positive integers or zero. How can I determine all possible solutions from this additional information. There are necessarily finite solutions as there are only finite positive integers for each x that are smaller than b. It is impossible to solve it by trying every possibility as the original system is way bigger (around 40 equations and 900 variables). I would appreciate every hint in the right direction.


There is a theory to solve systems of linear Diophantine equations over $\mathbb{Z}$, e.g., see here. Then one could select the nonneggative solutions.

In this particualr case, I find it helpful to first solve the system over the field of rational numbers. Then, by linear algebra, the system is equivalent to \begin{align*} x_1 & =\frac{b_1 + b_2 - b_3 - b_4 - b_5 + 2x_{10} + x_7 + 2x_8 + x_9}{2}, \\ x_2 & = \frac{b_1 - b_2 + b_3 - b_4 - b_5 + 2x_{10} + 2x_6 + x_7 + x_9}{2},\\ x_3 & = b_4 - x_{10} - x_6 - x_8, \\ x_4 & = b_5 - x_{10} - x_7 - x_9, \\ x_5 & = \frac{- b_1 + b_2 + b_3 + b_4 + b_5 - 2x_{10} - 2x_6 - 3x_7 - 2x_8 - x_9}{2}, \end{align*} where we can choose the variables on the LHS (namely $x_6,x_7,x_8,x_9,x_{10}$) as arbitrary rational numbers.

Now we can restrict the domain to nonnegative integers, and obtain the additional conditions on $x_6,\ldots ,x_{10}$: the expressions in the nominator for $x_1,x_2,x_5$ need to be even, and all expressions need to be nonnegative. There is still some work to do, however, but it seems easier to me than before. For example, if $b_4=b_5=0$, then necessarily $x_6=\ldots =x_{10}=0$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.