Exercise 6. We denote $[x]$ as the integer part of $x$. For example, $[15 / 4]=3$. We denote $f(n)=\left[\frac{n}{[\sqrt{n}]}\right]$. Find all integers $n$ such that $f(n+1)>f(n)$.
Solution to Exercise 6 Let $m=\left[\sqrt{n}\right]$. It is the unique integer such that $m^{2} \leqslant nf(n)$, we have $\frac{n+1}{m+1}>\frac{n}{m}$, hence $(n+1) m>n(m+1)$. After simplification, it follows that $m>n$. However, $m \leqslant \sqrt{n} \leqslant n$, so $nf(n)$ if and only if $\frac{n+1}{m}$ is an integer. Since $m^{2}<n+1<m^{2}+2 m+1$, this occurs if and only if $n+1=m^{2}+m$ or $n+1=m^{2}+2 m$. Conclusion: the solutions are the integers $n$ of the form $n=m^{2}+m-1$ or $n=m^{2}+2 m-1$.