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

Question

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.

Answer 1

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?

Answer 2

Note that if $n$ is even, say $n=2m$, this means we need $$(2m)^2-(2m) < m^2 \implies 3m^2 < 2m$$ No possible $m$.

If $n$ is odd, say $n=2m+1$, this means we need $$(2m+1)^2-(2m+1) < m^2 +m \implies 3m^2 + m < 0$$Again no possible $m$