13

I would like to prove that $$\big\lfloor \sqrt{n}+\sqrt{n+1}\big\rfloor = \big\lfloor \sqrt{n}+\sqrt{n+2}\big\rfloor$$ for all positive integers $n$.

Interestingly enough, this is not true for $\sqrt[3]{n}$ with a counter-example to be $n = 15$.

I wonder why that is true, and where in the proof it goes wrong for $\sqrt[3]{n}$.

MR_BD
  • 6,307
  • Possibly helpful: https://math.stackexchange.com/q/760330/42969 (and the linked questions), where it is proven that $\lfloor \sqrt{n}+\sqrt{n+1}\rfloor = \lfloor \sqrt{4n+1}\rfloor$. – Martin R Jun 05 '24 at 12:45
  • It can be shown that either $\big\lfloor \sqrt{n}+\sqrt{n+1}\big\rfloor = \big\lfloor \sqrt{n}+\sqrt{n+2}\big\rfloor$ or $\big\lfloor \sqrt{n}+\sqrt{n+1}\big\rfloor+1 = \big\lfloor \sqrt{n}+\sqrt{n+2}\big\rfloor$ holds for an integer $n$. – Sam Jun 05 '24 at 12:51
  • @Prem: No, it is not (verified with PARI/GP). – Martin R Jun 05 '24 at 13:01

3 Answers3

26

Suppose otherwise. Then there exists an integer $k$ such that $$\sqrt n+\sqrt{n+1}<k\le\sqrt{n}+\sqrt{n+2}$$ Squaring both sides: $$n+2\sqrt{n(n+1)}+(n+1)<k^2\le n+2\sqrt{n(n+2)}+(n+2)$$ $$2n+1+2\sqrt{n(n+1)}<k^2\le 2n+2+2\sqrt{n(n+2)}\tag{1}\label{eq1}$$ Now,$$\sqrt{n(n+1)}>n$$ and $$\sqrt{n(n+2)}<\sqrt{n^2+2n+1}=n+1$$ Putting these into $(1)$: $$2n+1+2n<k^2<2n+2+2(n+1)$$ $$4n+1<k^2<4n+4$$ So $k^2\equiv 2$ or $3\bmod4$, which is impossible.

TonyK
  • 68,059
  • That's a very clear proof. Thumbs up! – MR_BD Jun 05 '24 at 15:50
  • Computer calculations suggest that it is also true for the 4-th root. Does this method also work for that? I couldn't work it out. – MR_BD Jun 05 '24 at 16:21
  • 1
    Neither can I :-) – TonyK Jun 05 '24 at 16:26
  • 1
    @MR_BD yes, it can be made to work for 4th roots. By the same argument $(\sqrt[4] n+\sqrt[4]{n+1})^4>16n+1$, since expanding gives one term that is exactly $n$, one that is exactly $n+1$, and $14$ other terms that are in between. Because $\sqrt[4]{n}$ is concave, $\sqrt[4]n+\sqrt[4]{n+2}<2\sqrt[4]{n+1}=\sqrt[4]{16n+16}$. – Especially Lime Sep 19 '24 at 14:02
  • 1
    and since the only fourth powers mod $16$ are $0$ and $1$, you're done. (This last is what goes wrong for $\sqrt[3]{n}$: there are lots of cubes mod $8$.) – Especially Lime Sep 19 '24 at 14:05
  • 1
    It's not true for $\sqrt[8]{n}$, with $n=1525$ being a counterexample. – Especially Lime Sep 19 '24 at 14:11
  • @EspeciallyLime Thanks for the great comments. You can make it into an answer for better visibility – MR_BD Sep 19 '24 at 14:19
5

It could be proved using similar argument as proving $\lfloor \sqrt{n} + \sqrt{n+1} \rfloor = \lfloor \sqrt{4n+2} \rfloor$ in this post.

The reasoning goes like this: we first claim that $(\sqrt{n} + \sqrt{n+2})^2 \in (4n+3, 4n+4).$ Letting $y = \sqrt{n} + \sqrt{n+2}$, we will have $y \in (\sqrt{4n+3}, \sqrt{4n+4}).$ Thus $\lfloor y \rfloor$ should be an integer whose square is the first perfect square less than or equal to $4n+3$. But $4n+3$ is never a square. So $\lfloor y \rfloor = \lfloor \sqrt{4n+2} \rfloor$

Actually, both should be equal to $\lfloor \sqrt{4n+1} \rfloor$ also since $(4n+2)$ is never a perfect square.

For the $\sqrt[3]{n}$ case, I have not checked but I guess if you follow this logic, the bounding range might be larger and it might fit many perfect cubes in the range.

Pond-nj
  • 51
  • 3
0

Let consider $n$ going from $m^2$ to $m^2+2m$, then, since $\left(m+\frac12\right)^2=m^2+m+\frac14$, we need to check

  • for $n=m^2$ up to $n=m^2+m-1$

$$\big\lfloor \sqrt{m^2+m-1}+\sqrt{m^2+m}\big\rfloor =2m=\big\lfloor \sqrt{m^2+m-1}+\sqrt{m^2+m+1}\big\rfloor \tag 1$$

  • for $n=m^2+m$

$$\big\lfloor \sqrt{m^2+m}+\sqrt{m^2+m+1}\big\rfloor =2m+1=\big\lfloor \sqrt{m^2+m}+\sqrt{m^2+m+2}\big\rfloor \tag 2$$

  • for $n=m^2+2m$

$$\big\lfloor \sqrt{m^2+2m}+\sqrt{m^2+2m+1}\big\rfloor =2m+1=\big\lfloor \sqrt{m^2+2m}+\sqrt{m^2+2m+2}\big\rfloor \tag 3$$

Next we would have $n=m^2+2m+1=(m+1)^2$ which is covered by $(1)$ and so on.

Since $(1)$, $(2)$ and $(3)$ hold, the given identity holds.

For identities $(1)$, LHS is trivial and RHS is true by Bernoulli's inequality since

$$\sqrt{m^2+m-1}+\sqrt{m^2+m+1}<2m+1$$

Identities $(2)$ are true since for the LHS

$$\sqrt{m^2+m}+\sqrt{m^2+m+1}>2m+1$$

ideed by squaring both side

$$\sqrt{m^2+m}\sqrt{m^2+m+1}>m^2+m \iff \sqrt{m^2+m+1}>\sqrt{m^2+m}$$

For identities $(3)$, LHS is trivial and RHS is true by AM-GM

$$\sqrt{m^2+2m}+\sqrt{m^2+2m+2}<2m+2$$

$$\iff \sqrt{m^2+2m}\cdot \sqrt{m^2+2m+2}<m^2+2m+1$$

user
  • 162,563