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I am now trying to find proof for the following, which are significant to establishing proof for the Prime number relation that was originally stated in the question I posted here:

$$\Bigl \lfloor \frac{n+1}{\lfloor \sqrt{n+1} \rfloor}\Bigr\rfloor-\Bigl \lfloor \frac{n}{\lfloor \sqrt{n} \rfloor}\Bigr\rfloor \in {\{-1,0,1}\} \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad(A)$$

$$\frac{1}{2}\Biggl(\Biggl\lfloor \frac{(n+1)^2-1}{\Bigl\lfloor \sqrt{(n+1)^2-1} \Bigr\rfloor} \Biggr\rfloor-n\Biggr)-1=0\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad(B)$$

$$\Biggl\lfloor\frac{n+\bigl\lfloor(\frac{1}{2}n-\frac{1}{4n})^2\bigr\rfloor+1}{\bigl\lfloor \sqrt{n+\lfloor(\frac{1}{2}n-\frac{1}{4n})^2\rfloor+1} \bigr\rfloor}\Biggr\rfloor-\Biggl\lfloor\frac{n+\lfloor(\frac{1}{2}n-\frac{1}{4n})^2\rfloor}{\bigl\lfloor \sqrt{n+\bigl\lfloor(\frac{1}{2}n-\frac{1}{4n})^2\bigr\rfloor} \bigr\rfloor}\Biggr\rfloor-1=0\quad\quad\quad\quad(C)$$

$(B)$ and $(C)$ were conjectured from numerical observations of $(A)$

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(A) We divide in two cases:

Case 1: $n=m^2-1$ for some $m\in\mathbb N$.

In this case, we have that $$ \left\lfloor \dfrac{n+1}{\lfloor \sqrt{n+1} \rfloor} \right\rfloor-\left\lfloor \dfrac{n}{\lfloor \sqrt{n} \rfloor} \right\rfloor=\left\lfloor \dfrac{m^2}{\lfloor \sqrt{m^2} \rfloor} \right\rfloor-\left\lfloor \dfrac{m^2-1}{\lfloor \sqrt{m^2-1} \rfloor} \right\rfloor $$ and since $\lfloor \sqrt{m^2} \rfloor=m$ and $\lfloor \sqrt{m^2-1} \rfloor=m-1$, $$ \left\lfloor \dfrac{n+1}{\lfloor \sqrt{n+1} \rfloor} \right\rfloor-\left\lfloor \dfrac{n}{\lfloor \sqrt{n} \rfloor} \right\rfloor=\left\lfloor \dfrac{m^2}{m} \right\rfloor-\left\lfloor \dfrac{m^2-1}{m-1} \right\rfloor = m-(m+1)=-1 $$

Case 2: $n$ is not of the form $m^2-1$.

Pick $m\in\mathbb N$ such that $m^2\leq n \leq (m+1)^2-1$. Since $n\neq (m+1)^2-1$, we have $m^2\leq n \leq (m+1)^2-2$, and consequently, $$ m^2\leq n,\ n+1<(m+1)^2\ \Rightarrow\ m\leq \sqrt n,\ \sqrt{n+1}<m+1\ \Rightarrow\ \lfloor\sqrt n\rfloor=\lfloor\sqrt{n+1}\rfloor=m. $$ Then $$ \left|\frac{n+1}{\lfloor \sqrt{n+1}\rfloor}-\frac{n}{\lfloor \sqrt n\rfloor}\right|=\left|\frac{n+1}{m}-\frac{n}{m}\right|=\frac{1}{m}\leq 1, $$ and this implies that $$ \left\lfloor \dfrac{n+1}{\lfloor \sqrt{n+1} \rfloor} \right\rfloor-\left\lfloor \dfrac{n}{\lfloor \sqrt{n} \rfloor} \right\rfloor \in \{-1,0,1\}. $$

(B) We have that $\lfloor \sqrt{(n+1)^2-1}\rfloor=n$, because $$ n^2\leq(n+1)^2-1<(n+1)^2, $$ so $$ \frac{(n+1)^2-1}{\lfloor \sqrt{(n+1)^2-1}\rfloor} = \frac{n^2+2n}{n}=n+2, $$ from which follows the identity (B).

(C) First, we shall characterize the numbers such that $$ \left\lfloor \dfrac{n+1}{\lfloor \sqrt{n+1} \rfloor} \right\rfloor-\left\lfloor \dfrac{n}{\lfloor \sqrt{n} \rfloor} \right\rfloor=1. $$ By the Case 1 at the proof of (A), we already know that $n$ has not the form $m^2-1$. Pick $m\in\mathbb N$ such that $m^2\leq n\leq (m+1)^2-1$. By the Case 2 at the proof of (A), we know that $\lfloor\sqrt n\rfloor=\lfloor\sqrt{n+1}\rfloor=m$.

Put $n= m^2+k$, with $0\leq k \leq 2m$. Then $$ \dfrac{n+1}{\lfloor \sqrt{n+1} \rfloor} = m +\frac{k+1}{m}, $$ and $$ \dfrac{n}{\lfloor \sqrt{n} \rfloor} = m +\frac{k}{m}. $$ So, $$ \left\lfloor \dfrac{n+1}{\lfloor \sqrt{n+1} \rfloor} \right\rfloor-\left\lfloor \dfrac{n}{\lfloor \sqrt{n} \rfloor} \right\rfloor = \left\lfloor m+\dfrac{k+1}{m} \right\rfloor-\left\lfloor m+\dfrac{k}{m} \right\rfloor = \left\lfloor \dfrac{k+1}{m} \right\rfloor-\left\lfloor \dfrac{k}{m} \right\rfloor, $$ and the cases that this equals $1$ are precisely when $k=m-1$ and $k=2m-1$.

This way we proved that the only numbers $n$ that satisfiy $$ \left\lfloor \dfrac{n+1}{\lfloor \sqrt{n+1} \rfloor} \right\rfloor-\left\lfloor \dfrac{n}{\lfloor \sqrt{n} \rfloor} \right\rfloor=1, $$ Are precisely the $n$'s of the form $m^2+m-1$ or $m^2+2m-1$, for some natural $m$.

Now it only lasts to verify that the numbers $n+\left\lfloor\left(\dfrac{n}{2}-\dfrac{1}{4n}\right)^2\right\rfloor$ have one of these forms.

Let us do it. First, note that $$ \left(n-\frac{1}{2n}\right)^2=n^2-1+\frac{1}{4n^2}, $$ and since $0<\dfrac {1}{4n^2}<1$, $$ n^2-1<\left(n-\frac{1}{2n}\right)^2< n^2. $$ We divide in two cases:

Case 1. $n=2k$.

Then $$ \begin{array}{rcl} 4k^2-1<\left(n-\frac{1}{2n}\right)^2<4k^2 & \Rightarrow & k^2-1<k^2-\frac{1}{4}<\frac{1}{4}\left(n-\frac{1}{2n}\right)^2 < k^2 \\ & \Rightarrow & k^2-1<\left(\frac{n}{2}-\frac{1}{4n}\right)^2< k^2 \\ & \Rightarrow & \left\lfloor\left(\frac{n}{2}-\frac{1}{4n}\right)^2\right\rfloor = k^2-1 \\ & \Rightarrow & n+ \left\lfloor\left(\frac{n}{2}-\frac{1}{4n}\right)^2\right\rfloor = n + k^2 - 1 = k^2 + 2k -1. \end{array} $$ Case 2. $n=2k-1$.

Then $$ \begin{array}{rcl} 4k^2-4k<\left(n-\frac{1}{2n}\right)^2<4k^2 -4k+1 & \Rightarrow & k^2-k<\frac{1}{4}\left(n-\frac{1}{2n}\right)^2 < k^2 -k +\frac{1}{4} < k^2 -k +1\\ & \Rightarrow & k^2-k<\left(\frac{n}{2}-\frac{1}{4n}\right)^2< k^2 -k+1\\ & \Rightarrow & \left\lfloor\left(\frac{n}{2}-\frac{1}{4n}\right)^2\right\rfloor = k^2-k\\ & \Rightarrow & n+ \left\lfloor\left(\frac{n}{2}-\frac{1}{4n}\right)^2\right\rfloor = n + k^2 - k = k^2 +k-1. \end{array} $$

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  • $\begingroup$ haha well sure looks good so far wouldn't mind the proof for $\lfloor \sqrt{n+1}\rfloor = \lfloor \sqrt n\rfloor$ for $n$ not of the form $m^2-1$ while you are there anyway but sure I don't think I've got one for $\lfloor \sqrt{m^2-1} \rfloor=m-1$ either $\endgroup$ – Adam Oct 27 '18 at 23:13
  • $\begingroup$ Well I probably have one somewhere but still the one you've got after dinner would be a winner $\endgroup$ – Adam Oct 27 '18 at 23:18
  • $\begingroup$ Well, pick $m\in\mathbb N$ such that $m^2\leq n <(m+1)^2$. Then $m^2\leq n \leq(m+1)^2-1$. But since $n\neq (m+1)^2-1$, we have that $m^2\leq n \leq (m+1)^2-2$. Thus, both $n, n+1$ are numbers between $m^2$ and $(m+1)^2-1$... $\endgroup$ – André Porto Oct 27 '18 at 23:34
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    $\begingroup$ I incorporated the proof of the fact that $\lfloor\sqrt{n}\rfloor=\lfloor\sqrt{n+1}\rfloor$ $\endgroup$ – André Porto Oct 28 '18 at 0:04
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    $\begingroup$ So, just for completeness, I finished the proof of (C) at the answer. I hope it was useful. $\endgroup$ – André Porto Oct 28 '18 at 13:47

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