Q:

How to solve $n^2-n<\lfloor\frac{n^2}{4}\rfloor$ ,$n \in \mathbb{N}$?

How to solve this inequality?
$$n^2-n<\lfloor\frac{n^2}{4}\rfloor \\ n \in \mathbb{N}$$
I can solve normal inequalities but this includes floor function with which I am not familiar. and because of that I could do nothing from the beginning.

A:

HINT: It's always true that $\left\lfloor\frac{n^2}4\right\rfloor\le\frac{n^2}4$, so if $n^2-n<\left\lfloor\frac{n^2}4\right\rfloor$, then certainly $n^2-n<\frac{n^2}4$. What solutions does this inequality have in $\Bbb N$? Are any of them solutions to the original inequality?

