One can use common subexpressions to get a simple answer.  Note that a=n(n+3) must be a multiple of 4. Setting y=3x, we look for a+y is a multiple of 12 and a(a+y) is a multiple of 12y.  If we can pick y to meet a+y is a multiple of 12, then y is an integer and it suffices to also pick y being a divisor of a.  This is possible for n=5 and in general y being a 2 mod 3 divisor of n(n+3) should work.  I find x=(n+3)/3 works for many n.

Gerhard "Addition More Complex Than Multiplication?" Paseman, 2018.01.16.