Let $n$ be a positive integer, and define $f(n)$ as $n +\lfloor\sqrt{n}\rfloor$, where $\lfloor x\rfloor$ is the greatest positive integer less than or equal to $x$. Prove that the sequence $n, f(n), f(f(n)), f(f(f(n))), \ldots$ contains a perfect square.
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What are your thoughts, what efforts can you present to us here? – May 23 '17 at 13:13
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If $n = a^2-b$ with $b\le 2a$ then $\lfloor\sqrt{n}\rfloor = a-1$ and $f(n) = a^2-b+a-1$ – reuns May 23 '17 at 13:42
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We should look instead at $f(n) = n +2\lfloor\sqrt{n}\rfloor+1$ ? – reuns May 23 '17 at 13:46
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Does not seem to be true for $n=5$ ... $5,8,11,15,19,24,\cdots $ ... every second one seems to be one shy of the next square. – Donald Splutterwit May 23 '17 at 13:46
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interesting. @Do – G.H.Hardy May 23 '17 at 15:31
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@user1952009 can you please elaborate. – G.H.Hardy May 23 '17 at 15:32
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You need to look at $\displaystyle b = \min_a |n-a^2|$ – reuns May 23 '17 at 15:34
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@DonaldSplutterwit the sequence for $n=5$ is actually 5,7,9,... And 9 is a square. – fractal1729 May 23 '17 at 17:49
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@fractal1729 ... I was rounding up. Do I need to round down ? – Donald Splutterwit May 23 '17 at 17:52
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@DonaldSplutterwit Yeah, the floor function is for rounding down – fractal1729 May 23 '17 at 17:54
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@fractal1729 ... Thanks for that, it certainly seems to work with rounding down ... $n^2+n-1$ & $n^2+n$ will terminate at a square & cover every number along the way ... but I am not sure how to prove it. ... anyway dinner time ! – Donald Splutterwit May 23 '17 at 18:05
2 Answers
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Nice problem. Define $g(n)$ to be the remainder when $n$ is divided by $\lfloor\sqrt{n}\rfloor$. Also define $n_0 = n$, $n_{i+1}=f(n_i)$ for $i \geq 0$. Consider the sequence $g(n_0), g(n_1), g(n_2), \ldots$. We will prove that this sequence has infinitely many chunks of 0's separated by chunks of nonzero values. Note that if $g(n_i) > 0$ while $g(n_{i+1}) = 0$, that means $n_{i+1}$ is a perfect square. So we are essentially proving that the sequence $\{n_i\}$ has infinitely many perfect squares.
Let's set up notation first: suppose that $k^2 \leq n \leq (k+1)^2-1=k^2+2k$. Note that when we go from $n$ to $f(n)=n+k$, we either cross $(k+1)^2$ or we don't.
Case 1: If we don't, $n+k < (k+1)^2$ and $\lfloor\sqrt{n+k}\rfloor=\lfloor\sqrt{n}\rfloor=k$. Then it is clear that $g(n+k) = g(n)$.
Case 2: If we do, $n+k \geq (k+1)^2$ and $\lfloor\sqrt{n+k}\rfloor = \lfloor\sqrt{n}\rfloor+1 = k+1$. This means $k^2+2k \geq n \geq k^2+k+1$.
If $g(n) = 0$, then $k|n$, so $n = k^2+2k$. In this case $n+k = k^2+3k = (k+1)^2+(k-1)$, so $g(n+k) = k-1$.
If $g(n) > 0$ then $n = k^2+k+g(n)$. Thus $n+k = k^2+2k+g(n) = (k-1)^2+g(n)-1$, which means $g(n+k) = g(n)-1$.
Putting things together: So each time we step forward by one in the sequence $\{g(n_i)\}$, one of three things happens.
The value stays the same.
If the value is positive, value decreases by one.
If the value is zero, the value changes to some positive number.
Thus $\{g(n_i)\}$ will consist of chunks of zeros separated by chunks of nonzero values. So $\{n_i\}$ contains infinitely many perfect squares, and the proof is complete!
(Let me know in the comments if I should clarify any of these steps.)
fractal1729
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@TheGreatDuck The sequence is always increasing since $n$ is positive so $\lfloor\sqrt{n}\rfloor$ is also positive. Could you clarify what you mean by a loop? – fractal1729 May 24 '17 at 21:22
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1Sorry, I didn't realize you were referring to ${g(n_i)}$ instead of ${n_i}$. The reason that won't happen is that since ${n_i}$ is increasing, you will eventually cross a perfect square, which forces one of 2 or 3 to happen. – fractal1729 May 25 '17 at 01:29
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If $n$ is a square, you're done. If $n=m^2+k$ with $0\lt k\lt2m+1$, then in either one or two steps you'll be at a number of the form $(m+1)^2+k'$ with $0\le k'\lt k$. Induction (on $k$) now tells you that eventually you'll land on a square.
Barry Cipra
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