I'm trying to find possible solutions for the following equation:
$${ f\!\left(\matrix{x_1&x_2\\y_1+z_1&y_2+y_1z_1+z_2}\right) = f\!\left(\matrix{x_1&x_2\\y_1&y_2}\right) + f\!\left(\matrix{x_1&x_2\\z_1&z_2}\right) }$$
where the function ${ f }$ must satisty the anticommutative property with respect to the swapping of "upper" and "lower" arguments:
$${ f\!\left(\matrix{x_1&x_2\\y_1&y_2}\right)=-f\!\left(\matrix{y_1&y_2\\x_1&x_2}\right) }$$
(actually, it also has to be associative and have an identity since I want to construct it as a part of a certain binary operation, but let's put that aside for the moment)
Now, quite an interesting behavior emerges if we assume ${ z_1=x_1, z_2=x_2 }$ in the equation:
$${ f\!\left(\matrix{x_1&x_2\\y_1&y_2}\right) = f\!\left(\matrix{x_1&x_2\\y_1+x_1&y_2+y_1x_1+x_2}\right) }$$
(the term with only ${ x }$'s and ${ z }$'s in it gets zeroed out due to the anticommutative property)
If we fix the values of ${ x_1 }$ and ${ x_2 }$, the function becomes sort of periodic with respect to its lower arguments, but the curves of periodicity are parabolas rather than straight lines:
$${ f\!\left(\matrix{x_1&x_2\\y_1&y_2}\right)=f\!\left(\matrix{x_1&x_2\\y_1+nx_1&y_2+n(x_1y_1+x_2)+\frac{n(n-1)}{2}x_1^2}\right) }$$
for any given ${ y_1,y_2 }$ and for all ${ n∈\mathbb{Z} }$.
So, my question is: can you think of any nontrivial (explicitly dependent on all of the four arguments) solutions ${ f }$ to this equation that are maps ${ \mathbb{Z}→\mathbb{Z} }$ or ${ \mathbb{Q}→\mathbb{Q} }$? Solutions that involve other kinds of numbers are welcomed too but are a little out of the scope of my interest here.