1. Find all integers $n \geqslant 1$ such that $\lfloor\sqrt{n}\rfloor$ divides $n$.

Remark: The integer part of a real number $x \geqslant 0$ is the greatest integer less than or equal to $x$, and it is denoted by $\lfloor x\rfloor$. For example, $\lfloor 1.4\rfloor=1$, $\lfloor 2\rfloor=2$, and $\lfloor 2.9\rfloor=2$.

##
Solution

Let $k$ be an integer. We know that $k=\lfloor\sqrt{n}\rfloor$ if and only if $k \leqslant \sqrt{n}<k+1$, that is, if

$$
k^{2} \leqslant n<(k+1)^{2}=k^{2}+2 k+1
$$

or again $k^{2} \leqslant n \leqslant k^{2}+2 k$. Given this inequality, $k$ divides $n$ if and only if $n / k \in\{k, k+1, k+2\}$. At this point, we can proceed differently. Indeed, by considering $n-k^{2}$ and noting that it is a multiple of $k$, and satisfies $0 \leqslant n-k^{2} \leqslant 2 k$, we conclude that $n-k^{2} \in\{0, k, 2 k\}$.

In conclusion, the integers sought are those of the form $k^{2}, k(k+1)$ or $k(k+2)$, with $k \geqslant 1$.