Problem 3. Find all real numbers $a$ such that $4[a n]=n+[a[a n]]$ for any positive integer $n$ ( $[x]$ denotes the largest integer less than or equal to $x$ ).

Nikolai Nikolov
3. It follows by the given condition that $4(a n-1)$ $n+a(a n-1)-1$, i.e.,

$$
1+a^{2}-\frac{a+1}{n}<4 a<1+a^{2}+\frac{4}{n}
$$

Letting $n \rightarrow \infty$ gives $1+a^{2}=4 a$, so $a=2-\sqrt{3}$ or $a=2+\sqrt{3}$.

The given equality for $n=1$ yields that the first case is impossible.

In the second case set $b=\left[\frac{n}{a}\right]$ and $c=\frac{n}{a}-b$. Since $a=4-\frac{1}{a}$, then

$$
\begin{aligned}
n+[a[a n]] & =\left[n+a\left[4 n-\frac{n}{a}\right]\right]=[n+a(4 n-b-1)] \\
& =[a(4 n-1+c)]=\left[\left(4-\frac{1}{a}\right)(4 n-1+c)\right] \\
& =\left[4(4 n-1)-4\left(\frac{n}{a}-c\right)+\frac{1-c}{a}\right]=4(4 n-1-b) \\
& =4\left[4 n-\frac{n}{a}\right]=4[a n]
\end{aligned}
$$

Therefore $a=2+\sqrt{3}$ is the only solution of the problem.