Question

Find all real numbers $x$ such that
$[x^2+2x]=[x{]}^2+2[x]$ $($Here $\left[x\right]$ denotes the largest integer not exceeding $x.)$

Answer 1

Adding $1$ both sides, we get
$\left[\right(x+1{)}^2]=\left(\right[x+1]{)}^2$;
We have used $\left[x\right]+m=[x+m]$ for every integer $m$.
Suppose $x+1\le 0$. Then $[x+1]\le x+1\le 0$.
$\therefore \left(\right[x+1]{)}^2\ge (x+1{)}^2\ge \left[\right(x+1{)}^2]=([x+1]{)}^2$
Thus equality holds everywhere.
$\therefore [x+1]=x+1$ and thus $x+1$ is an integer using $x+1\le 0$,
we conclude that $x\in \{-1,-2,-3,....\}$.
Suppose $x+1>0$.
We have $(x+1{)}^2\ge [(x+1{)}^2]=\left(\right[x+1]{)}^2$
Moreover, we also have $(x+1{)}^2\le 1+[(x+1{)}^2]=1+\left(\right[x+1]{)}^2$
Thus we obtain $\left[x\right]+1=[x+1]\le (x+1)<\sqrt{1+\left(\right[x+1]{)}^2}=\sqrt{1+\left(\right[x]+1{)}^2}$
This shows that $x\in [n,\sqrt{1+(n+1{)}^2-1})$,
where $n\ge -1$ is an integer.
Thus the solution set is $\{-1,-2,-3,...\}\cup \left\{{\cup }_{n=-1}^{\infty }\right[n,\sqrt{1+(n+1{)}^2}-1\left)\right\}$
It is easy verify that all the real numbers in this set indeed satisfy the given equation.