Olimpiade Italiana di Matematica

Solution:

Let us first prove the case $n=1$. Knowing that $\left\lfloor\lambda^{2}\right\rfloor,\left\lfloor\lambda^{3}\right\rfloor$, and $\left\lfloor\lambda^{4}\right\rfloor$ are perfect squares, we want to show that $\lfloor\lambda\rfloor$ is a perfect square. Setting $a^{2}=\left\lfloor\lambda^{2}\right\rfloor$, we have the inequalities:
$$a^{2} \leq \lambda^{2}<a^{2}+1$$
from which, in particular, $\lambda^{2}<(a+1)^{2}$, hence $a \leq \lambda<a+1$ and $\lfloor\lambda\rfloor=a$. Squaring the previous chain of inequalities, we obtain
$$a^{4} \leq \lambda^{4}<\left(a^{2}+1\right)^{2}$$
and, since $a^{4}$ and $\left(a^{2}+1\right)^{2}$ are consecutive perfect squares, $\left\lfloor\lambda^{4}\right\rfloor=a^{4}$. Consequently $\lambda^{4}-a^{4}<1$. Now, using the hypothesis $\lambda \geq 1$,
$$\lambda^{3}-a^{3}=(\lambda-a)\left(\lambda^{2}+\lambda a+a^{2}\right) \leq(\lambda-a)\left(\lambda\left(\lambda^{2}+\lambda a+a^{2}\right)+a^{3}\right)=\lambda^{4}-a^{4}$$
so we also have
$$a^{3} \leq \lambda^{3}=a^{3}+\lambda^{3}-a^{3} \leq a^{3}+\lambda^{4}-a^{4}<a^{3}+1$$
which implies $\left\lfloor\lambda^{3}\right\rfloor=a^{3}$. By hypothesis, we therefore know that $a^{3}$ is a perfect square, hence $a=\lfloor\lambda\rfloor$ is also a perfect square, as required.

Let us now prove the general case, by induction on $n$. The base case, $n=1$, has been established. For the inductive step, suppose we know that, if $\left\lfloor\lambda^{n+1}\right\rfloor,\left\lfloor\lambda^{n+2}\right\rfloor, \ldots,\left\lfloor\lambda^{4 n}\right\rfloor$ are all perfect squares, then $\lfloor\lambda\rfloor$ is a perfect square. From the fact that $\left\lfloor\lambda^{(n+1)+1}\right\rfloor,\left\lfloor\lambda^{(n+1)+2}\right\rfloor, \ldots,\left\lfloor\lambda^{4(n+1)}\right\rfloor$ are all perfect squares, we want to deduce that $\lfloor\lambda\rfloor$ is a perfect square. If we also knew that $\left\lfloor\lambda^{n+1}\right\rfloor$ is a perfect square, then we could apply the inductive hypothesis and conclude immediately. Now, setting $\bar{\lambda}=\lambda^{n+1}$, in particular, $\left\lfloor\bar{\lambda}^{2}\right\rfloor,\left\lfloor\bar{\lambda}^{3}\right\rfloor$, and $\left\lfloor\bar{\lambda}^{4}\right\rfloor$ are three of our perfect squares. From the case $n=1$ applied to $\bar{\lambda}$ it therefore follows that $\left\lfloor\lambda^{n+1}\right\rfloor$ is a perfect square, and this concludes the proof.