Given that $a^2 = 9 + 4\sqrt{5}$, we need to: (a) Show that $a = \sqrt{5} + 2$. (b) Find the value of $a^4 - \frac{1}{a^4}$. (c) Show that $a^5 + \frac{1}{a^5} = 610\sqrt{5}$.

1. Problem Description

Given that $a^2 = 9 + 4\sqrt{5}$, we need to:

(a) Show that $a = \sqrt{5} + 2$.

(b) Find the value of $a^4 - \frac{1}{a^4}$.

(c) Show that $a^5 + \frac{1}{a^5} = 610\sqrt{5}$.

2. Solution Steps

(a) To show that $a = \sqrt{5} + 2$, we start with $a^2 = 9 + 4\sqrt{5}$. We can rewrite the right-hand side as a perfect square.

$a^2 = 9 + 4\sqrt{5} = 4 + 5 + 4\sqrt{5} = 2^2 + (\sqrt{5})^2 + 2 \cdot 2 \cdot \sqrt{5} = (2 + \sqrt{5})^2$

Taking the square root of both sides, we get:

$a = \sqrt{(2 + \sqrt{5})^2} = 2 + \sqrt{5} = \sqrt{5} + 2$ (since $a>0$)

(b) To find the value of $a^4 - \frac{1}{a^4}$, we first need to find $a^4$ and $\frac{1}{a^4}$.

Since $a = \sqrt{5} + 2$, we have $\frac{1}{a} = \frac{1}{\sqrt{5} + 2} = \frac{\sqrt{5} - 2}{(\sqrt{5} + 2)(\sqrt{5} - 2)} = \frac{\sqrt{5} - 2}{5 - 4} = \sqrt{5} - 2$.

Now, we find $a^2 = (\sqrt{5} + 2)^2 = 5 + 4\sqrt{5} + 4 = 9 + 4\sqrt{5}$.

Also, $\frac{1}{a^2} = (\frac{1}{a})^2 = (\sqrt{5} - 2)^2 = 5 - 4\sqrt{5} + 4 = 9 - 4\sqrt{5}$.

Next, $a^4 = (a^2)^2 = (9 + 4\sqrt{5})^2 = 81 + 72\sqrt{5} + 16 \cdot 5 = 81 + 72\sqrt{5} + 80 = 161 + 72\sqrt{5}$.

And $\frac{1}{a^4} = (\frac{1}{a^2})^2 = (9 - 4\sqrt{5})^2 = 81 - 72\sqrt{5} + 16 \cdot 5 = 81 - 72\sqrt{5} + 80 = 161 - 72\sqrt{5}$.

Therefore, $a^4 - \frac{1}{a^4} = (161 + 72\sqrt{5}) - (161 - 72\sqrt{5}) = 161 + 72\sqrt{5} - 161 + 72\sqrt{5} = 144\sqrt{5}$.

(c) To show that $a^5 + \frac{1}{a^5} = 610\sqrt{5}$, we can use the following identities:

$a^2 + \frac{1}{a^2} = 9 + 4\sqrt{5} + 9 - 4\sqrt{5} = 18$

$a + \frac{1}{a} = \sqrt{5} + 2 + \sqrt{5} - 2 = 2\sqrt{5}$

We have $a^3 + \frac{1}{a^3} = (a + \frac{1}{a})^3 - 3(a + \frac{1}{a}) = (2\sqrt{5})^3 - 3(2\sqrt{5}) = 8 \cdot 5\sqrt{5} - 6\sqrt{5} = 40\sqrt{5} - 6\sqrt{5} = 34\sqrt{5}$.

Also, $a^2 + \frac{1}{a^2} = 18$

$a^5 + \frac{1}{a^5} = (a^3 + \frac{1}{a^3})(a^2 + \frac{1}{a^2}) - (a + \frac{1}{a}) = (34\sqrt{5})(18) - (2\sqrt{5}) = 612\sqrt{5} - 2\sqrt{5} = 610\sqrt{5}$.

3. Final Answer

(a) $a = \sqrt{5} + 2$ is shown.

(b) $a^4 - \frac{1}{a^4} = 144\sqrt{5}$

(c) $a^5 + \frac{1}{a^5} = 610\sqrt{5}$ is shown.