Indian National Mathematical Olympiad

Solution:
Adding $1$ to 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

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

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