# For $n \in \mathbb{N}$ $\lfloor{\sqrt{n} + \sqrt{n+1}\rfloor} = \lfloor{\sqrt{4n+2}\rfloor}$

This is Exercise 3.20 from Apostol's book. Many of them seem tough and here is one of them which I am struggling with.

For $$n \in \mathbb{N}$$, prove that this identity is true: $$\Bigl\lfloor{\sqrt{n} + \sqrt{n+1}\Bigl\rfloor} = \Bigl\lfloor{\sqrt{4n+2}\Bigl\rfloor}$$

• is there any significance of the square brackets? Nov 23 '15 at 13:29
• @Chinny84: they usually signify "greatest integer less than contained value" or some similar wording... Nov 23 '15 at 13:31
• I assume that what you actually tried was $x=n^2+k$... Nov 23 '15 at 13:32
• The equation is not correct for general positive $x \in \mathbb{R}$. Try $x=\frac{1}{2}$ Nov 23 '15 at 13:43
• if it's the floor function, then it's not true. Check $1/2\leq x< 9/16$, for example. Similarly, the ceiling function also fails. Nov 23 '15 at 13:45

\begin{align*}(\sqrt{n} + \sqrt{n+1})^2 &= 4n+2 - 2\left(n+1/2 - \sqrt{n(n+1)}\right) \\ &= 4n+2 - 2(AM(n,n+1) - GM(n,n+1)) \\ &\in (4n+1, 4n+2).\end{align*}

(The first line is just algebra. In the second line, $AM$ and $GM$ are respectively the arithmetic and geometric means. To get the third line: $n < GM(n,n+1) < AM(n,n+1)=n+1/2$ by the AM-GM inequality, so $0 < AM(n,n+1)-GM(n,n+1) < 1/2$, and the third line follows.)

But there are no perfect squares between $4n+1$ and $4n+2$, and thus no integers between $\sqrt{n} + \sqrt{n+1}$ and $\sqrt{4n+2}$, QED.

• Nice work! Even i tried the first step which you have indicated.
– anonymous
Oct 5 '10 at 7:45
• But there are no perfect squares between $4n$ and $4n+2$ for $n≥1$ - are you sure? Oct 5 '10 at 7:52
• Crap, that was really stupid. Hopefully, it's correct now. Oct 5 '10 at 8:12

Note that $$\frac{1}{2} > \sqrt{n+\frac{1}{2}}-\sqrt{n} > \sqrt{n+1}- \sqrt{n+\frac{1}{2}} > 0,$$ since $f(x)=\sqrt{x+1/2}-\sqrt{x}$ is a decreasing function. The inequalities and the fact that $f(x)$ is decreasing follow from noting that $\sqrt{x+1/2} - \sqrt{x} = 1/2(\sqrt{x+1/2} + \sqrt{x}).$

So we can write $$1> \left( \sqrt{n+\frac{1}{2}}-\sqrt{n} \right) - \left( \sqrt{n+1}- \sqrt{n+\frac{1}{2}} \right) > 0.$$ Hence $$1 > 2\sqrt{n+\frac{1}{2}} - \left( \sqrt{n+1}+\sqrt{n} \right) > 0.$$ From which the result follows.

For clarity we've shown that we can write $$\sqrt{n+1}+\sqrt{n} + r = 2\sqrt{n+\frac{1}{2}} \quad \textrm{ for } 0 < r < 1,$$

where we note that we do not straddle an integer since $\sqrt{4n+1} < \sqrt{n} + \sqrt{n+1},$ and there are no integers between $\sqrt{4n+1}$ and $\sqrt{4n+2}.$

• Another interesting proof. Thanks
– anonymous
Oct 5 '10 at 12:44

The proof follows immediately from the following variant of the AM-GM inequality

$$\rm\quad\quad\quad\ 0 < n < m \ \Rightarrow\ 3\:n+m\ <\ (\sqrt{n}\ \ +\: \ \sqrt{m}\ )^2\ <\ 2\:n+2\:m\ \$$ (proof below)

Hence $$\rm\:\ m = n+1 \ \:\Rightarrow\ 4\:n+1\ \ <\ (\sqrt{n} + \sqrt{n+1})^2\ <\ 4\:n+2$$

The first inequality is easy to prove: expand middle term, then subtract $$\rm\ n+m\$$ from all the terms. Then it reduces on the left to $$\rm\ n\ < \sqrt{nm}\$$ via $$\rm\ n^2 < nm,\,$$ and the AM-GM inequality on the right.

Such minor variations on the $$\:$$ AM-GM $$\:$$ arise not too infrequently in practice (e.g. in competition problems).$$\:$$ Thus it is worthwhile to point them out in their full generality to help aid in recognizing them "in the wild".

First a kooky Lemma:

For $x > 1$ then $\sqrt{1 - \frac 1 x} + \sqrt{1 + \frac 3 x} > 2$

[$\sqrt{1 - \frac 1 x} + \sqrt{1 + \frac 3 x} > 2$ iff

$1 - \frac 1 x + 1 + \frac 3 x + 2\sqrt{(1 - \frac 1 x)(1 + \frac 3 x)} > 4$ iff

$2 + \frac 2 x + 2\sqrt{(1 + \frac 2 x -\frac 2 {x^2}} > 4$ iff

$1 + \frac 2 x -\frac 2 {x^2} > 1$ iff

$\frac 2 x -\frac 2 {x^2} > 0$ ]

================

Let $n^2 \le x < (n+1)^2$ or in other words $n^2 \le x \le n^2 + 2n$.

It's obvious:

$2n < \sqrt x + \sqrt {x + 1} < (n + 1) + (n + 1) = 2n + 2$.

So $\lfloor \sqrt x + \sqrt {x + 1} \rfloor = \{2n, 2n + 1\}$.

Likewise $2n = \sqrt{4n} < \sqrt{4x + 2} \le \sqrt{4n^2 + 8n + 2} < \sqrt{4^2 + 8 + 4} = 2(n + 1) = 2n + 2$.

So $\lfloor \sqrt {4x + 2} \rfloor = \{2n, 2n + 1\}$.

====================

Case 1:

$n^2 \le x < x + 1 \le n^2 + n < n^2 + n + \frac 1 4 = (n + \frac 1 2)^2$

Then $\sqrt x + \sqrt {x + 1} < n + \frac 1 2 + n + \frac 1 2 < 2n + 1$

So $\lfloor \sqrt x + \sqrt {x + 1} \rfloor = 2n$.

Likewise $\sqrt{4x + 2} \le \sqrt{4n^2 + 4(n -1) + 2} < \sqrt{4n^2 + 4n + 1} = 2(n + \frac 1 2) = 2n + 1$.

So $\lfloor \sqrt {4x + 2} \rfloor = 2n = \lfloor \sqrt x + \sqrt {x + 1} \rfloor$.

================

Case 2:

$n^2 + 2n + 1 \ge x + 1 > x \ge n^2 + n$.

Then $\sqrt{4x + 2} \ge \sqrt{4n^2 + 4n + 2} = 2(n + \frac 12) = 2n + 1$.

So $\lfloor \sqrt {4x + 2} \rfloor = 2n + 1$.

Now $\sqrt x + \sqrt {x + 1} \ge \sqrt{n^2 + n} + \sqrt{n^2 + n + 1} = \sqrt{n^2 + n + \frac 1 4 - \frac 1 4} + \sqrt{n^2 + n + \frac 1 4 + \frac 3 4}$

$= (n + \frac 1 2)[\sqrt{1 - \frac 1 {4(n + \frac 1 2)^2}} + \sqrt{1 + \frac 3 {4(n + \frac 1 2)^2}}]$

(remember the kooky lemma?)

$> (n + \frac 1 2)2 = 2n + 1$.

So $\lfloor \sqrt x + \sqrt {x + 1} \rfloor = 2n + 1 = \lfloor \sqrt{4x +2} \rfloor$.

• That's better :-) Nov 24 '15 at 3:19

Case 1: $n^2\le x\le(n+1)^2-2$ (for $n\ge 3$), then $$n\le \sqrt{x}\le\sqrt{(n+1)^2-2}<n+1, n<\sqrt{n^2+1}\le \sqrt{x+1}<\sqrt{(n+1)^2-1}=n+1$$ and $$2n<\sqrt{4n^2+2}\le \sqrt{4x+2}<\sqrt{4(n+1)^2-2}<2n+1$$ So $$2n<\sqrt{x}+\sqrt{x+1}<2n+2, \sqrt{4x+2}=2n$$ and $$2n+1\le[\sqrt{x}+\sqrt{x+1}]\le2n+2, [\sqrt{4x+2}]=2n+1.$$ Now we show $$\sqrt{x}+\sqrt{x+1}<2n+1.$$ Note $\sqrt{x}+\sqrt{x+1}$ is increasing and hence $$\sqrt{x}+\sqrt{x+1}\le\sqrt{(n+1)^2-2}+\sqrt{(n+1)^2-1}.$$ Note \begin{eqnarray} &&\sqrt{(n+1)^2-2}+\sqrt{(n+1)^2-1}-2n-1\\ &=&\frac{(n+1)^2-2-n^2}{\sqrt{(n+1)^2-2}+n^2}+\frac{(n+1)^2-1-n^2}{\sqrt{(n+1)^2-1}+n^2}-1\\ &<&\frac{4n-1}{\sqrt{(n+1)^2-2}+n^2}-1\\ &=&\frac{4n-1}{\sqrt{n^2+2n-1}+n^2}-1\\ &<&\frac{4n-1}{n+n^2}-1\\ &=&-\frac{n^2-3n+1}{n+n^2}\\ &<0 \end{eqnarray} for $n\ge 3$. Thus $$2n<\sqrt{(n+1)^2-2}+\sqrt{(n+1)^2-1}<2n+1$$ or $$[\sqrt{x}+\sqrt{x+1}=2n+1$$ Case 2: $x=(n+1)^2-1$. \begin{eqnarray} &&\sqrt{x}+\sqrt{x+1}\\ &=&\sqrt{(n+1)^2-1}+n+1\\ &<&2(n+1) \end{eqnarray} and \begin{eqnarray} &&\sqrt{x}+\sqrt{x+1}\\ &=&\sqrt{(n+1)^2-1}+n+1\\ &>&2n+1 \end{eqnarray} and hence $$[\sqrt{x}+\sqrt{x+1}]=2(n+1).$$ On the other hand $$\sqrt{4x+2}=\sqrt{4(n+1)^2-2}<2(n+1)$$ and $$\sqrt{4x+2}=\sqrt{4(n+1)^2-2}>2n+1$$ and hence $$[\sqrt{4x+2}]=2n+2$$ In both cases, $$[\sqrt{x}+\sqrt{x+1}]=[\sqrt{4x+2}].$$ Case 3: for $n=1,2$, it is easy to check the inequality is true.

Taking the listed approach, we get

$$[\sqrt{n^2+k}+\sqrt{n^2+k+1}]=[\sqrt{4n^2+4k +2}]$$

As $k$ relates to $n$, we have a floor-function boundary for the RHS when $k\in\{-1,0\}$ and also at $k\in\{n-1,n\}$, so these values need to be checked on the LHS to verify that these are consecutive boundaries there as well. For $k\in\{-1,0\}$, the LHS shows the same behavior as the RHS, so we only need to show that the LHS also has a boundary at $k\in\{n-1,n\}$ by comparing with $2n+1$ from the RHS:

$$\left[\sqrt{n^2+n-1}+\sqrt{n^2+n}\right]\, R\, 2n+1 \\ \text{ vs. }\\ \left[\sqrt{n^2+n}+\sqrt{n^2+n+1}\right]\, R\, 2n+1$$

Squaring yields

$$\left[2n^2+2n-1+2\sqrt{(n^2+n)(n^2+n-1)}\right]\, R\, 4n^2+4n+1\\ \left[2\sqrt{(n^2+n)(n^2+n-1)}\right]\, R\, 2n^2+2n+2$$

and it should be clear that $[\sqrt{n^2+n-1}+\sqrt{n^2+n}]\lt 2n+1$. Next we have

$$\left[2n^2+2n+1+2\sqrt{(n^2+n)(n^2+n+1)}\right]\, R\, 4n^2+4n+1\\ \left[2\sqrt{(n^2+n)(n^2+n+1)}\right]\, R\, 2n^2+2n$$

and it should also be clear that $[\sqrt{n^2+n}+\sqrt{n^2+n+1}]\ge 2n+1$, completing the proof.

Here's a more pedestrian solution that does not rely on any number-theoretic result. We show the inequalities $$\lfloor \sqrt k +\sqrt{k+1} \rfloor \leq \sqrt{4k+2} < \lfloor \sqrt k +\sqrt{k+1} \rfloor+1.$$

Note that $$\sqrt k + \sqrt{k+1}\leq \sqrt{4k+2} \iff 2\sqrt k \sqrt{k+1} \leq 2k+1 \iff 0\leq 1$$, hence $$\lfloor \sqrt k +\sqrt{k+1} \rfloor \leq \sqrt{4k+2}.$$

For the remaining strict inequality, note by squaring and comparing integers that

\begin{align} \sqrt{4k+2} < \lfloor \sqrt k +\sqrt{k+1}\rfloor + 1 &\iff 4k+2 < \lfloor \sqrt k +\sqrt{k+1}\rfloor^2 + 2\lfloor \sqrt k +\sqrt{k+1}\rfloor + 1 \\ &\iff 4k+3 \leq \lfloor \sqrt k +\sqrt{k+1}\rfloor^2 + 2\lfloor \sqrt k +\sqrt{k+1}\rfloor + 1 \\ &\iff 4k+2 \leq \lfloor \sqrt k +\sqrt{k+1}\rfloor^2 + 2\lfloor \sqrt k +\sqrt{k+1}\rfloor. \end{align}

The inequality is true for $$k\in \{0,1\}$$ so we assume in what follows that $$k\geq 2$$. Since $$\lfloor z \rfloor > z-1$$ and $$\sqrt k +\sqrt{k+1}-1\geq 2\sqrt k -1\geq 0$$,we have

\begin{align} \lfloor \sqrt k +\sqrt{k+1}\rfloor^2 + 2\lfloor \sqrt k +\sqrt{k+1}\rfloor &\geq (\sqrt k +\sqrt{k+1}-1)^2 + 2(\sqrt k +\sqrt{k+1}-1)\\ &\geq (2\sqrt k -1)^2 + 2(\sqrt k +\sqrt{k+1}-1)\\ &=4k + 1 + 2(\sqrt{k+1}-1)\\ &\geq 4k + 1 + 2(\sqrt{3}-1)\\ &\geq 4k+2. \end{align}