Floor & floor again

Case(1): Consider : $x$ is a positive real.Take : $x=I+f$ where : $I,f$ are integer and fractional part of : $x$ respectively.

Putting this in the expression it becomes : $\lfloor x \lfloor x \rfloor \rfloor \\ =\lfloor (I+f)I \rfloor \\=\lfloor I^2 +If \rfloor \\=I^2 + \lfloor If \rfloor$

As : $0 \leq f <1$ so : $0 \leq If$ .So for each : $I$ the fractional part can be subdivided into : $I$ parts.An example is suppose : $I=2$ so : $If$ becomes : $2f$ .If we subdivide two part in : $f$ i.e. : $0 \leq f < 0.5$ and : $0.5 \leq f <1$ then two different integers will generate .here this will be : $2^2+0,2^2+1$ .The integers that can be expressed in the form of : $\lfloor x \lfloor x \rfloor \rfloor$ are in the form of : $I^2,I^2+1,I^2+2,....I^2+I$ .So maximum : $I$ will be : $\lfloor \sqrt{2016} \rfloor=44$ .

The solutions will be : $(1^2+0);(2^2+0,2^2+1);(3^2+0,3^2+1,3^2+2);,........;(44^2+0,44^2+1,,,,,,44^2+43)$ .Total number of solutions are : $=1+2+3+4+.....+44=990$

Case(2): Consider : $x$ is a negative real number .Take : $x=-I-f$ where : $I$ ,: $f > 0$ are integer and fractional part of : $x$ respectively

Putting this in the expression it becomes : $\lfloor x \lfloor x \rfloor \rfloor \\=\lfloor -(I+f)\times -(I+1) \rfloor \\=\lfloor (I+f)(I+1) \rfloor \\=\lfloor I^2+I + f(I+1) \rfloor \\=I^2+I + \lfloor f(I+1) \rfloor$ .

Here : $f$ is multiplied with : $I+1$ . so we can subdivide : $f$ into : $I+1$ parts each will give an integer separately.AS example if we take : $I=2$ then : $3f$ will give three integers by subdividing : $f$ as : $0 \lt f<1/3; 1/3 \leq f <2/3 ;2/3 \leq f <1$ .So, for any integer : $I$ in the given range it will also give total : $I+1$ integers as solutions.The maximum value of : $I=44$ .The maximum solution will be : $=44^2 +44 +44=2024 >2016$ .So : $8$ solutions should be discarded.

The solution set is : $(1^2+1+0,1^2+1+1);(2^2+2+0,2^2+2+1,2^2+2+2);(3^2+3+0,3^2+3+1,3^2+3+2,3^2+3+3);.........,(44^2+44+0,44^2+44+1,....44^2+44+36)$ .

Total number of solutions are : $(2+3+4+5+......+45)-8=1026$ .

Total solutions are : $1026+990=2016$

I think every number from 1 to 2016 can be expressed.