There is no unique solution to the system.
You essentially have two equations in three unknowns:
C + D &= A\\
B - 2C &= 0.
Given a value of $A$, if you pick any value for $D$, then this will give you a value for $C$, which in turn will give you a value for $B$. Or you can pick any value for $C$, then solve for $D$ and $B$. (You can also pick a value for $B$, so long as it is even, and get a value for $C$ and $D$).
Even if you require all values to be positive integers, there is still no unique solution: $D$ gets restricted to only the values $1$, $2,\ldots,A-1$, but each of those values will give you a valid solution with $C$ and $B$ positive integers; so the only case where you would have a unique solution in positive integers is if $A=2$ (in which case, $B=2$ and $C=D=1$).