# Question #30720

Jun 18, 2016

Let me try

#### Explanation:

Let
$\sqrt{n + 1} + \sqrt{n - 1} \text{ is rational and can be expressed by } \frac{p}{q}$
where p and q prime to each other and$\text{ } q \ne 0$

So
$\sqrt{n + 1} + \sqrt{n - 1} = \frac{p}{q} \ldots \ldots \ldots \left(1\right)$

Inverting (1) we get

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

$\implies \frac{\sqrt{n + 1} - \sqrt{n - 1}}{\left(\sqrt{n + 1} + \sqrt{n - 1}\right) \left(\sqrt{n + 1} - \sqrt{n - 1}\right)} = \frac{q}{p}$

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

$\implies \left(\sqrt{n + 1} - \sqrt{n - 1}\right) = \frac{2 q}{p} \ldots . . \left(2\right)$

Adding (1) and (2) we get

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

$\implies \sqrt{n + 1} = \frac{{p}^{2} + 2 {q}^{2}}{2 p q} \ldots . . \left(3\right)$

Similarly subtracting (2) from (1) we get

$\implies \sqrt{n - 1} = \frac{{p}^{2} - 2 {q}^{2}}{2 p q} \ldots . . \left(4\right)$

Since p and q are integers then eqution (3) and equation(4) reveal

that both$\text{ } \sqrt{n + 1} \mathmr{and} \sqrt{n - 1}$

$\textcolor{b l u e}{\text{ are rational as their RHS rational}}$

So both $\left(n + 1\right) \mathmr{and} \left(n - 1\right) \text{ will be perfect square}$

Their difference becomes $\left(n + 1\right) - \left(n - 1\right) = 2$

But we know any two perfect square differ by at least by 3

Hence it can be inferred that there is no positive integer for which

$\sqrt{n + 1} + \sqrt{n - 1} \text{ is rational}$