Exercise 4.1.28* ($\left \lfloor \sqrt{n} + \sqrt{n+1} \right \rfloor = \left \lfloor \sqrt{n} + \sqrt{n+2} \right \rfloor.$)

Proof.

(a)

Suppose for the sake of contradiction that $\lfloor \sqrt{n}+\sqrt{n+1}\rfloor \ne \lfloor \sqrt{n}+\sqrt{n+2}\rfloor$.

Then there exits a positive integer $k$ such that

$$\begin{array}{lll}\hfill \sqrt{n}+\sqrt{n+1}<k\le \sqrt{n}+\sqrt{n+2}.& & \hfill \text{(1)}\end{array}$$

(Take $k=\lfloor \sqrt{n}+\sqrt{n+2}\rfloor$. Then $k\le \sqrt{n}+\sqrt{n+2}$. If $k\le \sqrt{n}+\sqrt{n+1}$, then

$$k\le \sqrt{n}+\sqrt{n+1}<\sqrt{n}+\sqrt{n+2}<k+1,$$

so $k=\lfloor \sqrt{n}+\sqrt{n+1}\rfloor =\lfloor \sqrt{n}+\sqrt{n+2}\rfloor$, and this contradicts our hypothesis. Therefore $\sqrt{n}+\sqrt{n+1}<k$).

Put $r=k^2-4n$, so that $r$ is an integer. By (1), $2\sqrt{n}<\sqrt{n}+\sqrt{n+1}<k$, thus $r=k^2-4n>0$, so $r\ge 1$.

We use this expression of $k=\sqrt{4n+r}$ in inequalities (1):

$$\begin{array}{lll}\hfill \sqrt{n}+\sqrt{n+1}<\sqrt{4n+r}\le \sqrt{n}+\sqrt{n+2}.& & \hfill \text{(2)}\end{array}$$

These inequalities are equivalent to

$$\begin{array}{llllll}\hfill \sqrt{n+1}& <\sqrt{4n+r}-\sqrt{n}\le \sqrt{n+2},& \hfill & & \hfill & \\ \hfill n+1& <5n+r-2\sqrt{n(4n+r)}\le n+2,& \hfill & & \hfill & \\ \hfill -4n-r+1& <-2\sqrt{n(4n+r)}\le -4n-r+2,& \hfill & & \hfill & \\ \hfill 4n+r-2& \le 2\sqrt{n(4n+r)}<4n+r-1,& \hfill & & \hfill & \\ \hfill 16n^2+r^2+4+8\mathit{nr}-16n-4r& \le 16n^2+4\mathit{nr}<16n^2+r^2+1+8\mathit{nr}-8n-2r,& \hfill & & \hfill & \\ \hfill r^2+4-16n-4r& \le -4\mathit{nr}<r^2+1-8n-2r,& \hfill & & \hfill & \\ \hfill -16n+{(r-2)}^2& \le -4\mathit{nr}<-8n+{(r-1)}^2,& \hfill & & \hfill & \\ \hfill 8n-{(r-1)}^2& <4\mathit{nr}\le 16n-{(r-2)}^2.& \hfill (3)& & \hfill & & \hfill \end{array}$$

Since $4\mathit{nr}\le 16n-{(r-2)}^2\le 16n$, we obtain $r\le 4$.

• If $r=4$, then $4\mathit{nr}=16n>16n-{(r-2)}^2=16n-4$. This contradicts (3), therefore $r\ne 4$.
• If $r=1$, then $8n-{(r-1)}^2=8n>4n$. This contradicts (3), therefore $r\ne 1$.
• If $r=2$, then $k^2=4n+2$, so $k^2\equiv 2(\mathit{mod}4)$. This congruence has no solution, therefore $r\ne 2$.
• If $r=3$, then $k^2=4n+3$, so $k^2\equiv 3(\mathit{mod}4)$. This congruence has no solution, therefore $r\ne 3$.

This shows that the inequalities (4) are not satisfied for any pair of positive integers $n,r$. Equivalently, the inequalities (1) are not satisfied for any pair of positive integers $n,k$. Therefore, for any positive integer $n$,

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

(b)

If $n=15$, $$\begin{array}{lll}\hfill \sqrt[3]{15}+\sqrt[3]{16}<4.99& & \hfill \\ \hfill \sqrt[3]{15}+\sqrt[3]{17}>5.03.& & \hfill \end{array}$$

Therefore

$$\lfloor \sqrt[3]{15}+\sqrt[3]{16}\rfloor =4\ne 5=\lfloor \sqrt[3]{15}+\sqrt[3]{17}\rfloor$$

(we obtain similar inequalities for n = 42, 274, 421, …).

So

is not true for every integer $n$, □