$\lfloor \sqrt n+\sqrt {n+1}+\sqrt{n+2}+\sqrt{n+3}+\sqrt{n+4}\rfloor=\lfloor\sqrt {25n+49}\rfloor$ is true?

I found the following relational expression by using computer:

For any natural number $n$, $$\lfloor \sqrt n+\sqrt {n+1}+\sqrt{n+2}+\sqrt{n+3}+\sqrt{n+4}\rfloor=\lfloor\sqrt {25n+49}\rfloor.$$

Note that $\lfloor x\rfloor$ is the largest integer not greater than $x$.

I can neither prove this nor find any counterexample even by using computer. Could you show me how to prove this? Or could you get an counterexample?

Update: I've just asked a related question.

Generalization of $\lfloor \sqrt n+\sqrt {n+1}+\sqrt{n+2}+\sqrt{n+3}+\sqrt{n+4}\rfloor=\lfloor\sqrt {25n+49}\rfloor$

Note that by the strict concavity of $\sqrt{x}$, Jensen's Inequality says $$\hspace{-1cm}\sqrt{n}+\sqrt{n+1}+\sqrt{n+2}+\sqrt{n+3}+\sqrt{n+4}\lt5\sqrt{n+2}=\sqrt{25n+50}\tag{1}$$ More precisely, Taylor's Formula with remainder says \begin{align} \sqrt{n+2+x} =\sqrt{n+2}+\frac{x}{2\sqrt{n+2}}-\int_0^x\frac{(x-t)\,\mathrm{d}t}{4\sqrt{n+2+t}^3}\tag{2} \end{align} Summing $(2)$ for $x\in\{-2,-1,0,1,2\}$ yields $$\hspace{-1cm}5\sqrt{n+2}-\frac5{4n^{3/2}}\lt\sqrt{n}+\sqrt{n+1}+\sqrt{n+2}+\sqrt{n+3}+\sqrt{n+4}\lt5\sqrt{n+2}\tag{3}$$ Since \begin{align} \sqrt{25n+50}-\sqrt{25n+49} &=\frac1{\sqrt{25n+50}+\sqrt{25n+49}}\\ &\gt\frac1{2\sqrt{25n+50}}\\ &\gt\frac5{4n^{3/2}}\quad\text{for }n\ge14\tag{4} \end{align} we get that for $n\ge14$, $$\hspace{-5mm}\sqrt{25n+49}\lt\sqrt{n}+\sqrt{n+1}+\sqrt{n+2}+\sqrt{n+3}+\sqrt{n+4}\lt\sqrt{25n+50}\tag{5}$$ $(5)$ says that for $n\ge14$, $$\left\lfloor\sqrt{n}+\sqrt{n+1}+\sqrt{n+2}+\sqrt{n+3}+\sqrt{n+4}\right\rfloor =\left\lfloor\sqrt{25n+49}\right\rfloor\tag{6}$$ It is simple to verify $(6)$ for $1\le n\lt14$ (it is false for $n=0$).

• Why does (5) lead to (6)? – nbubis Aug 27 '13 at 11:41
• @nbubis: If $25n+50$ is not a perfect square then $\left\lfloor\sqrt{25n+49}\right\rfloor=\left\lfloor\sqrt{25n+50}\right\rfloor$. If $25n+50$ is a perfect square, then both $\left\lfloor\sqrt{n}+\sqrt{n+1}+\sqrt{n+2}+\sqrt{n+3}+\sqrt{n+4}\right\rfloor$ and $\left\lfloor\sqrt{25n+49}\right\rfloor$ are $\sqrt{25n+50}-1$. – robjohn Aug 27 '13 at 12:01
• @robjohn: Thank you very much for great proof! – mathlove Aug 27 '13 at 14:35
• one can avoid Jensen's Inequality by checking $\sqrt{x}+\sqrt{x+4} \lt 2 \sqrt{x+2}$ and $\sqrt{x+1}+\sqrt{x+3} \lt 2 \sqrt{x+2}$ by squaring – miracle173 Aug 27 '13 at 16:03
• @miracle173: why would one want to avoid such a useful inequality? In any case, I didn't actually use Jensen, I mention it at the start for support, but use Taylor's Theorem which gives an upper and lower bound. – robjohn Aug 27 '13 at 16:08

This isn't a real proof, but note that: $$\sqrt{25n+49}=5\sqrt{n+2-1/25}\sim5\sqrt{n+2}$$ Which is quite close to the sum: $$\sum_{k+0}^4 \sqrt{n+k}$$ So as $n$ becomes larger the difference becomes smaller.

If you look at the series expansion around $n\to\infty$ of: $$f(n)=\sqrt{n}+\sqrt{n+1}+\sqrt{n+2}+\sqrt{n+3}+\sqrt{n+4}-\sqrt{25 n+49}$$ $$\sim\frac{1}{10n^{1/2}}+O\left(n^{-3/2}\right)$$ So it's clear that the only problem could arise if one side is equal to $M+\epsilon$, for some very small $\epsilon < 0.1n^{-1/2}$.

I would try and establish a lower bound for $\epsilon$ and try to arrive at a contradiction.

Let $$L(n):= \sqrt{n-2}+\sqrt{n-1}+\sqrt{n}+\sqrt{n+1}+\sqrt{n+2},\quad R(n):=\sqrt{25n-1}\ .$$ We have to show that $\lfloor L(n)\rfloor=\lfloor R(n)\rfloor$ for all $n\geq2$.

The function $$f(x):=\sqrt{1+x}+\sqrt{1-x}$$ has $f(0)=2$, $\>f'(0)=0$, and $$f''(x)=-{1\over4}\bigl((1+x)^{-3/2}+(1-x)^{-3/2}\bigr)\doteq-{1\over2}\quad\bigl(|x|\ll1\bigr)\ .$$ It follows that $$f(x)\doteq2-{x^2\over4}\quad\bigl(|x|\ll1\bigr)\ .$$ We therefore have $$L(n)=\sqrt{n}\left(1+f({1\over n})+f({2\over n})\right)\doteq\sqrt{n}\bigl(5-{5\over4n^2}\bigr)\doteq\sqrt{25n-{25\over 2n}}\ .\tag{1}$$ Assume now that $$n=k^2+\ell, \qquad 0\leq\ell<2k+1\ .$$ Then one can easily conclude from $(1)$ (with $R(n)$ its even simpler) that $$\lfloor L(n)\rfloor=\lfloor R(n)\rfloor=\cases{5k-1\quad&(\ell=0)\cr 5k&(\ell\geq1)\cr}\quad.$$

I want to proof this only by using basic arithmetic and therefore avoiding Jensen's Inequality and Taylor's Theorem.

It's simpler if one replaces $n$ by $n-2$. On then gets

$$\lfloor \sqrt{n-2}+\sqrt{n-1}+\sqrt n+\sqrt {n+1}+\sqrt{n+2}\rfloor=\lfloor\sqrt {25n-1}\rfloor$$

It is necessary to show that

$$\sqrt {25n-1} \lt \sqrt{n-2}+\sqrt{n-1}+\sqrt n+\sqrt {n+1}+\sqrt{n+2} \tag{1}$$

$$\sqrt{n-2}+\sqrt{n-1}+\sqrt n+\sqrt {n+1}+\sqrt{n+2} \lt \sqrt {25n} \tag{2}$$

We first prove

$$\sqrt{x-b}+\sqrt{x+b} \lt \sqrt{x-a}+\sqrt{x+a} , 0 \lt a \lt b \lt x \tag{3}$$

by squaring $(3)$ we get

$$\sqrt{x-b}\sqrt{x+b} \lt \sqrt{x-a}\sqrt{x+a}$$

and by squaring this again we get

$$x^2-b^2 \lt x^2-a^2$$

and therefore

$$a^2 \lt b^2$$

The arguments is also valid in the opposite direction , because all numbers we squared where positive.

$$\sqrt{n-2}+\sqrt{n-1}+\sqrt n+\sqrt {n+1}+\sqrt{n+2} \lt 5 \sqrt{n}$$ follows immediately from $$\sqrt{n-2}+\sqrt{n+2}=\sqrt{n}+\sqrt{n}$$ which can be deduced from $(3)$ and $b=2$ and $a=0$, and from $$\sqrt{n-1}+\sqrt{n+1}=\sqrt{n}+\sqrt{n}$$ which can be deduced from $(3)$ and $b=1$ and $a=0$,

So $(2)$ is proofen.

To proof $(1)$ we notice that

$$\sqrt{n-2}+\sqrt{n-1}+\sqrt n+\sqrt {n+1}+\sqrt{n+2} \gt 2\sqrt{n-2}+\sqrt n+ 2 \sqrt{n+2}$$

because $$\sqrt{n-1}+ \sqrt{n+1} \gt \sqrt{n-2}+ \sqrt{n+2}$$

So it is sufficiont to check if

$$2 \sqrt{n-2}+\sqrt n+ 2 \sqrt{n+2} \gt \sqrt{25n-1}$$

We square

$$2 \sqrt{n-2} + 2 \sqrt{n+2} \gt \sqrt{25n-1} - \sqrt{n}$$

and get

$$8\,\sqrt{n-2}\,\sqrt{n+2}+8\,n>-2\,\sqrt{n}\,\sqrt{25\,n-1}+26\,n-1$$ and bring the squareroot terms to the LHS and the non non-squareroot terms to the RHS of the inequation

$$2\,\sqrt{n}\,\sqrt{25\,n-1}+8\,\sqrt{n-2}\,\sqrt{n+2} \gt 18\, n-1$$

We repeate this process until we get

$$9216\,n^3-185664\,n^2+20544\,n-66049 \gt 0$$

The polynomial can be written as

$$\left(n-21\right)\,\left(9216\,n^2+7872\,n+185856\right)+3836927$$

so for $n \ge 21$ the polynomial is positive and $(1)$ is valid.

• Maybe not. Just because $\sqrt{25n-1} < \sqrt{}+...$ diesn't mean that $\lfloor\sqrt{25n-1}\rfloor = \lfloor\sqrt{}+...\rfloor$. – marty cohen Oct 13 '13 at 18:08
• @marty cohen: but $\sqrt{25n-1} \lt \sqrt{}+... \lt \sqrt{25n}$ does mean that $\lfloor\sqrt{25n-1}\rfloor = \lfloor\sqrt{}+...\rfloor$. This is $(1)$ and $(2)$. There is no integer in $(\sqrt{25n-1},\sqrt{25n})$ because there is none in $(25n-1,25n)$. – miracle173 Oct 14 '13 at 5:24