f(1)=21=a+b+c+d; d=21-(a+b+c)
f(21)=9261a+441b+21c+d=31061;
9261a+441b+21c+21-(a+b+c)=31061;
9260a+440b+20c=31040;
926a+44b+2c=3104.
463a+22b+c=1552.
a<4, so:
let a=3: 1389+22b+c=1552, 22b+c=163.
If c has to be positive, then b<8. If b cannot be 7, then let b=6:
132+c=163, c=163-132=31.
So, if a=3, b=6, c=31 and d=21-(a+b+c)=21-40=-19.
If c can be negative, let b=8, 176+c=163, c=-13.
So, if a=3, b=8, c=-13, then d=21-(a+b+c)=21-(-2)=23.
The question doesn't specify any other constraints so there are several solutions, including negative values.