Croatian Mathematical Competitions

If $n = p$ is a prime number, then $l = p$ so it is obvious that $l > \frac{n}{4}$. For $k$ we have: $kp = (y - 1)(y + 1)$, so
$$\begin{cases} y - 1 = k_1, & y + 1 = k_2p, \\ y - 1 = k_1p, & y + 1 = k_2. \end{cases} \quad \text{or}$$
In each case, adding these two equations gives us $2y > p$. If $4k \le p$ (i.e. $4k < p$ since $p$ is prime), then it would be $y^2 = kp + 1 < \frac{p^2}{4} + 1$, i.e. $p^2 < 4y^2 < p^2 + 4$. This is not possible since there are no perfect squares between $p^2$ and $p^2 + 4$.
We will prove that there does not exist an odd composite number $n > 3$ such that $4l > n$ and $4k > n$. Assume on the contrary that such $n$ exists. We can write $n$ in the form
$$n = x^2 \cdot p_1 \cdots p_s$$
with $p_1, \dots, p_s$ different prime numbers. Clearly, $l = p_1 \cdots p_s$ and we have
$$4p_1 \cdots p_s = 4l > n = x^2p_1 \cdots p_s.$$
Hence $x^2 < 4$, so $x = 1$, i.e.
$$n = p_1 \cdots p_s.$$
Let $y$ denote the smallest number greater than $1$ such that $y^2 - 1$ is divisible by $n$ (such $y$ exists since $(n+1)^2 - 1 = n(n+2)$ is divisible by $n$). Clearly $kn + 1 = y^2$, and it is easy to see that $k > \frac{n}{4} \iff 2y > n$.
Now we can write $n$ in the form $n = pr$, where $p = p_i$ for some $i$, and $r$ is the product of all $p_j$ for $j \neq i$. Since $n$ is a composite number it follows that $r > 1$.
By the Chinese remainder theorem there exists a unique $T$, $0 \le T < n$, such that
$$T \equiv 1 \pmod{r},$$
$$T \equiv -1 \pmod{p}.$$
Now look at the number $S = n - T$. We have:
$$S \equiv -1 \pmod{r},$$
$$S \equiv 1 \pmod{p},$$
so $T^2 \equiv S^2 \equiv 1 \pmod{n}$. Therefore, both $T$ and $S$ are candidates for $y$. Clearly, one of them is less than $\frac{n}{2}$, so $k < \frac{n}{4}$, which is a contradiction.