An algebra problem by Puneet Pinku

First note that for $\sqrt{x - 1}$ to be real we require that $x \ge 1$ . But this implies that $\sqrt{x + 1} \ge \sqrt{2} \gt 1$ , and thus there are $\boxed{0}$ real numbers $x$ that satisfy the equation.

Note that if we were to have tried squaring both sides we would have found that

$x + 1 + (x - 1) + 2\sqrt{(x + 1)(x - 1)} = 1 \Longrightarrow 4(x^{2} - 1) = (1 - 2x)^{2} \Longrightarrow -4 = 1 - 4x \Longrightarrow x = \dfrac{5}{4}$ .

But with this value of $x$ we find that $\sqrt{x + 1} + \sqrt{x - 1} = 2$ , i.e., in this case the "square both sides" approach yields just one potential solution which we must discard.

$\text{If } \sqrt[]{n-1} - \sqrt[]{n+1} \text{ is rational, } \\ \sqrt[]{n-1} - \sqrt[]{n+1} = \frac{a}{b} ; (a, b) = 1, b \neq 0; a, b \in Q \text{ ... } \fbox{1}\\ \implies \dfrac{1}{\sqrt[]{n-1} - \sqrt[]{n+1}} = \frac{b}{a} \\ \implies \frac{b}{a} = \dfrac{\sqrt[]{n-1} + \sqrt[]{n+1}}{(\sqrt[]{n-1} - \sqrt[]{n+1})(\sqrt[]{n-1} - \sqrt[]{n+1})} \\ \implies \frac{b}{a} = \dfrac{\sqrt[]{n-1} + \sqrt[]{n+1}}{(n+1) - (n-1)} \\ \implies \frac{b}{a} = \dfrac{\sqrt[]{n-1} + \sqrt[]{n+1}}{2} \\ \implies \sqrt[]{n-1} + \sqrt[]{n+1} = \dfrac{2b}{a} \text{ ... } \fbox{2} \\ \\ \fbox{1} + \fbox{2} \\ \implies \frac{a}{b} + \frac{2b}{a} = \sqrt[]{n-1} - \sqrt[]{n+1} + \sqrt[]{n-1} - \sqrt[]{n+1} \\ \implies \dfrac{a^2+2b^2}{2ab} = 2\sqrt[]{n+1} \text{ ... } \fbox{3} \\ \fbox{1} - \fbox{2} \\ \implies \frac{a}{b} - \frac{2b}{a} = \sqrt[]{n-1} - \sqrt[]{n+1} - (\sqrt[]{n-1} - \sqrt[]{n+1}) \\ \implies \dfrac{a^2-2b^2}{2ab} = \sqrt[]{n-1} \text{ ... } \fbox{4} \\ \\ \text{From } \fbox{3} \text{ and } \fbox{4} \\ \sqrt[]{n-1} \text{ and } \sqrt[]{n+1} \text{ are rational, as 'a' and 'b' are rational.}\\ \implies n-1 \text{ and } n+1 \text{ are perfect squares} \\ \text{but, there is no possibility for a square no to differ by 2. Therefore, there is no real solutions for that equation making it rational.}$

http://brilliant.com/problems/rationality-2

$\begin{aligned} \sqrt{x+1}+\sqrt{x-1}&=2\\ \dfrac {2}{\sqrt{x+1}-\sqrt{x-1}}&=2\\ \sqrt{x+1}-\sqrt{x-1}=1\\ \sqrt{x+1}=\frac{3}{2}\\ \end{aligned}$

But this causes $\sqrt{x+1}>1$ , so there are no solutions.