
3. Find all real numbers $x$ such that

$$
\left[x^{2}+2 x\right]=[x]^{2}+2[x]
$$

(Here $[x]$ denotes the largest integer not exceeding $x$.)


## Solution:

Adding 1 both sides, the equation reduces to

$$
\left[(x+1)^{2}\right]=([x+1])^{2}
$$

we have used $[x]+m=[x+m]$ for every integer $m$. Suppose $x+1 \leq 0$. Then $[x+1] \leq$ $x+1 \leq 0$. Thus

$$
([x+1])^{2} \geq(x+1)^{2} \geq\left[(x+1)^{2}\right]=([x+1])^{2}
$$

Thus equality holds everywhere. This gives $[x+1]=x+1$ and thus $x+1$ is an integer. Using $x+1 \leq 0$, we conclude that

$$
x \in\{-1,-2,-3, \ldots\}
$$

Suppose $x+1>0$. We have

$$
(x+1)^{2} \geq\left[(x+1)^{2}\right]=([x+1])^{2}
$$

Moreover, we also have

$$
(x+1)^{2} \leq 1+\left[(x+1)^{2}\right]=1+([x+1])^{2}
$$

Thus we obtain

$$
[x]+1=[x+1] \leq(x+1)<\sqrt{1+([x+1])^{2}}=\sqrt{1+([x]+1)^{2}}
$$

This shows that

$$
x \in\left[n, \sqrt{1+(n+1)^{2}}-1\right)
$$

where $n \geq-1$ is an integer. Thus the solution set is

$$
\{-1,-2,-3, \ldots\} \cup\left\{\cup_{n=-1}^{\infty}\left[n, \sqrt{1+(n+1)^{2}}-1\right)\right\}
$$

It is easy verify that all the real numbers in this set indeed satisfy the given equation.
