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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.

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  • $\begingroup$ What does that little mark on the start of the last line mean? $\endgroup$ – fleablood Dec 13 '18 at 18:09
  • $\begingroup$ I guess "prove that" $\endgroup$ – Federico Dec 13 '18 at 18:09
  • $\begingroup$ Experimenting, it seems that $f(2^k+k-2)=2^{k-1}$ $\endgroup$ – Federico Dec 13 '18 at 18:10
  • $\begingroup$ Yeah. It means prove that $\endgroup$ – furfur Dec 13 '18 at 18:15
  • $\begingroup$ The answer says that f(2^k + k - 2)=(2^(k -1))^2 $\endgroup$ – furfur Dec 13 '18 at 18:17
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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 $n_k=2^k+k-2$.

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