Prove $\sqrt{n+1}+\sqrt{n-1}$ is irrational for all $n\in\Bbb Z^+$

Question

Question:- Prove that there exists no positive integer $n$ for which $\sqrt {n-1}+\sqrt{n+1}$ is a rational number.

Answer 1

Prove by contradiction. Let $a,b\in\Bbb Z^+$.

$$\sqrt{n+1}+\sqrt{n-1}=\frac{a}{b}\,\stackrel{2}\implies\, 2n+2\sqrt{n^2-1}=\frac{a^2}{b^2}$$

$$\implies n^2-1=\left(\frac{\frac{a^2}{b^2}-2n}{2}\right)^2\implies n=\pm 1,$$

because $n^2-1$ is a square of a rational (thus of an integer), and the only consecutive integer squares are $0,1$.

Get a contradiction. I assume you know why $\sqrt{2}$ is irrational (see here for why $\sqrt{n}$ with $n\in\Bbb Z_{\ge 0}$ is either an integer or irrational).

Answer 2

Assuming $\sqrt{n-1}+\sqrt{n+1}\in\mathbb{Q}$, then the square also belong to $\mathbb{Q}$, so $\sqrt{n^2-1}$ belongs to $\mathbb{Q}$, so $n^2-1$ is a square, so $n=1$. But for $n=1$ we have $\sqrt{n-1}+\sqrt{n+1}=\sqrt{2}\not\in\mathbb{Q}$.

Answer 3

Alternatively, let $p = \sqrt{n-1}, q = \sqrt{n+1} \Rightarrow n = \dfrac{(n-1)+(n+1)}{2}=\dfrac{p^2+q^2}{2}= \dfrac{1}{2}\cdot \left((p+q)^2 - 2pq\right)\Rightarrow n + pq = \dfrac{(p+q)^2}{2}\Rightarrow n + \sqrt{n^2-1} = \dfrac{(p+q)^2}{2}\Rightarrow \sqrt{n^2-1} = \dfrac{(p+q)^2}{2}-n \in \mathbb{Q}$ since $p+q \in \mathbb{Q}$. Thus $n^2-1 = k^2 \Rightarrow n^2-k^2 = 1 \Rightarrow (n-k)(n+k) = 1 \Rightarrow n - k = n+k = 1 \Rightarrow n = 1, k = 0 \Rightarrow p+q = \sqrt{2} \in \mathbb{Q}$, contradiction.

Answer 4

Let $\sqrt{n+1}+\sqrt{n-1}=t$; then $$ t=\sqrt{n+1}+\sqrt{n-1}\frac{\sqrt{n+1}-\sqrt{n-1}}{\sqrt{n+1}-\sqrt{n-1}} =\frac{2}{\sqrt{n+1}-\sqrt{n-1}} $$ and so $$ \sqrt{n+1}-\sqrt{n-1}=\frac{2}{t} $$ Thus $$ 2\sqrt{n+1}=\frac{2}{t}+t=\frac{2+t^2}{t} $$ and so $$ n+1=\frac{4+4t^2+t^4}{4t^2} $$ and we can rearrange this to $$ t^4-4nt^2+4=0 $$ A rational root of this polynomial must be in the set $\{1,-1,2,-2,4,-4\}$, but none of these numbers is a root: \begin{align} &(\pm1)^4-4n(\pm1)^2+4=5-4n\ne0\\ &(\pm2)^4-4n(\pm2)^2+4=20-16n=4(5-4n)\ne0\\ &(\pm4)^4-4n(\pm4)^2+4=260-64n=4(65-16n)\ne0 \end{align}