# Question #f992a

Jul 4, 2016

The equation has infinitely many solutions. See below.

#### Explanation:

The important thing to realize in order to solve this problem is that if square roots are multiplied, you can group them together. In math terms:
$\sqrt{a} \cdot \sqrt{b} = \sqrt{a b}$

So, for example, $\sqrt{32} \cdot \sqrt{2} = \sqrt{32 \cdot 2} = \sqrt{64} = 8$.

In this equation, we have $\sqrt{{n}^{2} - 1} \cdot \sqrt{{n}^{2} + 1}$. Using the rule described above, we can combine them into one square root:
$\sqrt{{n}^{2} - 1} \cdot \sqrt{{n}^{2} + 1} = \sqrt{\left({n}^{2} - 1\right) \left({n}^{2} + 1\right)}$

The expression $\left({n}^{2} - 1\right) \left({n}^{2} + 1\right)$ may not look like anything we know at first, but recall the difference of squares property:
$\left(a - b\right) \left(a + b\right) = {a}^{2} - {b}^{2}$

$\left({n}^{2} - 1\right) \left({n}^{2} + 1\right)$ is actually a difference of squares, with $a = {n}^{2}$ and $b = 1$. Since this simplifies into ${a}^{2} - {b}^{2}$, we can say:
$\left({n}^{2} - 1\right) \left({n}^{2} + 1\right) = {\left({n}^{2}\right)}^{2} - {\left(1\right)}^{2} = {n}^{4} - 1$

We've just simplified $\sqrt{{n}^{2} - 1} \cdot \sqrt{{n}^{2} + 1}$ into $\sqrt{{n}^{4} - 1}$. Our problem now looks like:
$\sqrt{{n}^{4} - 1} = \sqrt{{n}^{4} - 1}$

Hm...both sides are the same! What does that mean? It means the equation has infinitely many solutions! Any value of $n$ will satisfy this equation.