IMO Team Selection Test 2

We first prove the statement for $n=1$. Write $x=a+r$, with $a\ge 1$ an integer and $0\le r<1$. Suppose $\lfloor x^2\rfloor ,\lfloor x^3\rfloor$ and $\lfloor x^4\rfloor$ are squares. Then we have $a\le x<a+1$, from which it follows that $a^2\le x^2<(a+1{)}^2$. Hence, $x^2$ is squeezed between two consecutive squares. However, $\lfloor x^2\rfloor$ is a square, hence the only possibility is that $\lfloor x^2\rfloor =a^2$. We conclude that $a^2\le x^2<a^2+1$.

Completely analogously, we also get that $(a^2{)}^2\le x^4<(a^2+1{)}^2$ and hence $\lfloor x^4\rfloor =a^4$. We conclude that $a^4\le x^4<a^4+1$.

Moreover, we have that $x^3\ge a^3$. Now suppose that $x^3\ge a^3+1$, i.e. $$x^4\ge x(a^3+1)=(a+r)(a^3+1)=a^4+r{a}^3+a+r\ge a^4+a\ge a^4+1,$$ which gives a contradiction. Hence, $x^3<a^3+1$, which yields that $\lfloor x^3\rfloor =a^3$. This is also a square, hence $a$ must be a square itself. We see that $\lfloor x\rfloor$ is a square.

Now we will finish the proof with induction to $n$. The induction basis has just been proved. Now let $k\ge 1$ and suppose that the statement is proved for $n=k$. Consider a real number $x\ge 1$ with the property that $\lfloor x^{k+2}\rfloor ,\lfloor x^{k+3}\rfloor ,\dots ,\lfloor x^{4k+4}\rfloor$ are all squares. In particular, $\lfloor x^{2(k+1)}\rfloor ,\lfloor x^{3(k+1)}\rfloor$, and $\lfloor x^{4(k+1)}\rfloor$ are all squares. We can now apply the case $n=1$ on $x^{k+1}$ (which is a real number greater than or equal to 1) and find that $\lfloor x^{k+1}\rfloor$ is also a square. Now we know that $\lfloor x^{k+1}\rfloor ,\lfloor x^{k+2}\rfloor ,\dots ,\lfloor x^{4k}\rfloor$ are all squares and using the induction hypothesis, we obtain that $\lfloor x\rfloor$ is a square as well. This completes the proof by induction. $\square$