Show that there is no positive …

Let us assume that there is a positive integer n for {tex}\sqrt{n-1}+\sqrt{n+1}{/tex}which is rational and equal to {tex}\frac pq{/tex}, where p and q are positive integers and (q {tex}\neq{/tex} 0).
{tex}\sqrt { n - 1 } + \sqrt { n + 1 } = \frac { p } { q }{/tex}......(i)
or, {tex}\frac { q } { p } = \frac { 1 } { \sqrt { n - 1 } + \sqrt { n + 1 } }{/tex}

on multiplication of numerator and denominator by {tex}\sqrt{n-1}-\sqrt{n+1}{/tex} we get
{tex}= \frac { \sqrt { n - 1 } - \sqrt { n + 1 } } { ( \sqrt { n - 1 } + \sqrt { n + 1 } ) ( \sqrt { n - 1 } - \sqrt { n + 1 } ) }{/tex}
{tex}= \frac { \sqrt { n - 1 } - \sqrt { n + 1 } } { ( n - 1 ) - ( n + 1 ) } = \frac { \sqrt { n - 1 } - \sqrt { n + 1 } } { - 2 }{/tex}
or, {tex}\sqrt { n + 1 } - \sqrt { n - 1 } = \frac { 2 q } { p }{/tex} ........(ii)
On adding (i) and (ii), we get
{tex}2 \sqrt { n + 1 } = \frac { p } { q } + \frac { 2 q } { p } = \frac { p ^ { 2 } + 2 q ^ { 2 } } { p q }{/tex}

{tex}\sqrt{n+1}\;=\frac{p^2+2q^2}{2pq}{/tex}...............(iii)

From (i) and (ii),
{tex}\style{font-family:Arial}{\sqrt{n-1}\;=\frac{p^2-2q^2}{2pq}}{/tex}........(iv)
In RHS of (iii) and (iv) {tex}\frac{p^2+2q^2}{2pq}\;and\;\frac{\displaystyle p^2-2q^2}{\displaystyle2pq}\;are\;rational\;number\;because\;p\;and\;q\;are\;positive\;integers{/tex}

But it is possible only when (n + 1) and (n - 1) both are perfect squares.

Now n+1-(n-1)=n+1-n+1=2

Hence they differ by 2 and two perfect squares never differ by 2.

So both (n + 1) and (n -1 ) cannot be perfect squares.   

Hence there is no positive integer n for which {tex}\style{font-family:Arial}{\sqrt{n-1\;}+\sqrt{n+1}}{/tex} is rational