So this is probably not really a puzzle. As I've been exploring "extremes" of KenKen space, it is fairly obvious to me that 2 region KenKen are plentiful but almost always uninteresting. In an n x n Latin Square, you need to only specify the right T(n-1) squares, T(n) being the triangular number 1+2+...+n, to specify the whole Latin Square. Well, this isn't that hard with smallest or largest sums/products. What was interesting to me was to split the two regions to have the same value, which this and only this puzzle in arithmetic seems to be able to do (with region redrawing obviously possible).