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

Solution 1:

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

(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.

Solution 2:

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 $\sqrt{4n+2}.$

Solution 3:

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".

Solution 4:

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

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

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So now to answer your question:

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

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

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