It's Only Algebra

This is a rip-off from Calvin Lin's problem, which he gives a full explanation of the math behind this

Summing Up Certain Roots of Unity

The only possible totals are $(0, 2, 4, 5, 6, 8, 10)$ , or $7$ in all

More direct solutions are welcome, it'd be interesting to see the different ways of solving this

Edit: To simplify solving this problem, consider the transform $a=1-a'$ , and do the same for all of the variables. The $-1$ drops out in all four equations, and we divide everything by $-1$ to end up where we started from. So that means that given any solution, there exist a complement solution where for every $1$ , there is now a $0$ , and vice versa. So, for example, if we proved that there exists no solution such that the total is $3$ , then there exists no solution such that the total is $7$ . It's very easy to show that there can't be any solutions such that the total is $1$ , so that means there is no solution such that the total is $9$ .

As an example of where the total can be $5$ for a solution (which means that its complement is also a solution), we have, using the exact same four equations as displayed, but with variables replaced by integers, we have, letting $(a, c, d, g, j)=(1,1,1,1,1)$

$\begin{cases}1+0-0-1=0\\ 0+1-1-0=0\\ 1+0-0-1=0\\ 0-1+1-0-1+0-0+1=0\end{cases}$

It is not hard to show that the total can be any of the even numbers, $(0, 2, 4, 6, 8, 10)$ , after which we're left with proving that the total cannot be $3$ , which implies it cannot be $7$ either. To prove the case for even totals, the four equations can be rearranged as follows

$\begin{cases}(a-f)+(b-g)=0\\ (b-g)+(c-h)=0\\ (d-i)+(e-j)=0\\ (b-g)-(c-h)+(d-i)-(e-j)=0\end{cases}$

and from this we can show that

$(a-f)=-(b-g)=(c-h)=(d-i)=-(e-j)$

which means if we set this value at $0$ , we can flip any pair of variables arbitrarily to create another solution, thus either raising or lowering the total by $2$ s. Hence, the total can be of any even number from $0$ to $10$ .

Finally, we'll look at the special case where the total is $3$ . Consider the original set of equations, and start with assuming that all of the variables are $0$ which is a solution.

$\begin{cases}a+b-f-g=0\\ b+c-g-h=0\\ d+e-i-j=0\\ b-c+d-e-g+h-i+j=0\end{cases}$

For the total to be $3$ , we have to flip $3$ of the variables from $0$ to $1$ without affecting the sum of of each equation which is $0$ . Clearly, it is not possible that an odd number of variables can be flipped in any one equation---either none or two can be flipped. A little analysis of the equations given shows this to be impossible. For example, if $a$ and $f$ are flipped, since there are no instances of $a$ and $f$ elsewhere, then the 3rd flipped variable will change the sum of $0$ of any equation it appears.

Hence, no solutions exist for which the total is $3$ , and consequently $7$ as well. And so we have the final list of possible totals $(0, 2, 4, 5, 6, 8, 10)$ , or 7 distinct totals.