Mathematica competitions in Croatia

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 $$\{\begin{array}{ll}y-1=k_1,& y+1=k_{2}p,\\ y-1=k_{1}p,& y+1=k_2.\end{array}\text{or}$$ 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 $$4p_1\cdots p_s=4l>n=x^2{p}_1\cdots p_s.$$ 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 $k>\frac{n}{4}⟺2y>n$. 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 $j\ne i$. 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 $$T\equiv 1(\mathrm{mod}r),$$ $$T\equiv -1(\mathrm{mod}p).$$ Now look at the number $S=n-T$. We have: $$S\equiv -1(\mathrm{mod}r),$$ $$S\equiv 1(\mathrm{mod}p),$$ so $T^2\equiv S^2\equiv 1(\mathrm{mod}n)$. 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.