Half of my 2-digit number N
is also a 2-digit number, one of its digit being the sum of N's digit, the other their difference.
All 4 digits are different.

By the way, I notice that the algebraic approach demonstrates that there is only one solution. But that approach never took advantage of all 4 digits being different. Therefore, the fact that all 4 digits are different is not needed to solve the problem.