# Number of elements in the range of $f(x) = \lfloor x^2 \rfloor - \lfloor x \rfloor^2$

If $$f(x) = \lfloor x^2 \rfloor - \lfloor x \rfloor^2$$, and $$x \in [0,n]$$ where $$n \in \mathbb{Z}$$ then the number of elements in the range of $$f(x)$$ is:-

$$1.) \ 2n+1$$ $$2.)\ 4n-3$$ $$3.)\ 3n-3$$ $$4.)\ 2n-1$$

I tried it like this.

Let $$x=I+f$$, where $$I$$ is $$\lfloor x \rfloor$$ and $$f$$ is fractional part of $$x$$. So,

$$f(x) = \lfloor (I+f)^2 \rfloor - I^2$$ $$f(x) = \require{cancel}\cancel{I^2} +\lfloor 2If+f^2 \rfloor - \cancel{I^2}$$ $$f(x) =\lfloor 2If+f^2 \rfloor$$

But I am stuck on this step, please help me out. Thanks in advance.

• $0\leq f(x) < 2(n-1) +1 +1 =2n$ Apr 15, 2019 at 15:36
• The answer is 2n-1
– user585765
Apr 15, 2019 at 15:37

It is less helpful to try to solve for $$f(x)$$ explicitly though, as you can already figure out an upper and lower bound for $$f(x)$$; $$l \le x < l+1$$ without having to do that.
Let us write $$l = \lfloor x \rfloor$$. Then $$f(x) = \lfloor x^2 \rfloor - l^2$$ is always an integer, and can be as large as $$(l+1)^2-1-l^2 = 2l$$ [for $$x$$ close to $$l+1$$] but no larger; and of course can be as small as 0, if $$x-l^2$$ is close to 0 [make sure you see why]. So the number of elements in the range of $$\lfloor x^2 \rfloor - l^2$$ $$=$$ $$\lfloor x^2 \rfloor - \lfloor x \rfloor ^2$$ $$=f(x)$$; where $$x$$ varies $$l \le x < l+1$$ can be precisely as large as $$2l+1$$. [make sure you see why]. So the number of elements in the range of $$\lfloor x^2 \rfloor - l^2$$ $$=$$ $$\lfloor x^2 \rfloor - \lfloor x \rfloor ^2$$ $$=f(x)$$; where $$x$$ varies $$l \le x \le l+1$$ can be precisely as large as $$2l+1$$. [make sure you see why]
So can you use this to conclude that the number of elements in the range of $$\lfloor x^2 \rfloor - \lfloor x \rfloor ^2$$; $$x \in [0,n]$$ can be as large as $$2(n-1)+1$$ $$=2n-1$$ but no larger.