Question

Show that there is no positive integer n for which $\sqrt{n-1}+\sqrt{n+1}$ is rational.

Answer 1

Suppose there exists a positive integer $n$ for which is a rational number.
Where p and q positive integers and $q\ne 0$

$\Rightarrow \frac{q}{p}=\frac{1}{\sqrt{n-1}+\sqrt{n+1}}$

$\Rightarrow \frac{q}{p}=\frac{\sqrt{n-1}-\sqrt{n+1}}{(n-1)-(n+1)}=\frac{\sqrt{n-1}-\sqrt{n+1}}{-2}$

$\Rightarrow \frac{2q}{p}=\sqrt{n+1}-\sqrt{n-1}$

$(\sqrt{n-1}+\sqrt{n+1})+(\sqrt{n+1}-\sqrt{n-1})=\frac{p}{q}+\frac{2q}{p}$

$\Rightarrow 2\sqrt{n+1}=\frac{p^2+2q^2}{pq}$

$\Rightarrow \sqrt{n+1}=\frac{p^2+2q^2}{2pq}$.....(1)

$(\sqrt{n-1}+\sqrt{n+1})-(\sqrt{n+1}-\sqrt{n-1})=\frac{p}{q}-\frac{2q}{p}$

$\Rightarrow 2\sqrt{n-1}=\frac{p^2-2q^2}{pq}$

$\Rightarrow \sqrt{n-1}=\frac{p^2-2q^2}{2pq}$.....(2)

From eq1 and eq 2

$\sqrt{n+1}and\sqrt{n-1}arerational$ $\because pandqareintegers\Rightarrow \frac{p^2+2q^2}{2pq}and\frac{p^2-2q^2}{2pq}$

$\Rightarrow n+1andn-1perfectsquareofpositiveintegers$.

Now $\sqrt{(}n+1)-\sqrt{(}n-1)=2$ which is not possible since any two perfect squares differ by at least $3$.

Hence there is no positive integer $n$ for which is a rational number.