A problem with functions defined on positive integers.

Question

Where [x] denotes the greatest integer number, which does not exceed x.

I need some help please. The proof should also be at high school level. Please don’t use hard or complex things.

Answer 1

Hint. Show that if $f(n)=N^2$ then for $0\leq k\leq N$ $$\begin{align} &f(n+2k+1)=(N+k)^2+N-k, &\lfloor\sqrt{f(n+2k+1)}\rfloor=N+k,\\ &f(n+2k+2)=(N+k)^2+2N, &\lfloor\sqrt{f(n+2k+2)}\rfloor=N+k. \end{align}$$ Therefore, in this range, we have a perfect square only for $k=N$, i.e. $f(n+2N+1)=(2N)^2$. Hence $$f(1)=1^2\to f(4)=2^2\to f(9)=4^2\to f(18)=8^2\to f(35)=16^2.$$ We may conclude that $f(n)$ is a perfect square if $n$ belongs to the sequence $(n_k)_{k\geq 1}=1,4,9,18,35,\dots$ , that is $n_k=2^k+k-2$.