# Thread: Problem #19: Exponent Solution

1. ## Problem #19: Exponent Solution

Here's a more straightforward problem for College Algebra level.

Let a > 1 be an integer. Find all integers x, such that

$\left ( \sqrt{ a + \sqrt{a^2 - 1} } \right ) ^x + \left ( \sqrt{ a - \sqrt{a^2 - 1} } \right ) ^x = 2a$

-Dan

2. ## Re: Problem #19: Exponent Solution

Hint:
Spoiler:

Let $y = \sqrt{a + \sqrt{a^2 - 1}}$. What is 1/y?

-Dan

3. ## Re: Problem #19: Exponent Solution

No takers this time? Here's my solution:

Spoiler:

$\left ( \sqrt{ a + \sqrt{a^2 - 1} } \right ) ^x + \left ( \sqrt{ a - \sqrt{a^2 - 1} } \right ) ^x = 2a$

Let $y = \sqrt{ a + \sqrt{a^2 - 1} }$

Then
$\frac{1}{y} = \frac{1}{\sqrt{ a + \sqrt{a^2 - 1} }} = \frac{1}{\sqrt{ a + \sqrt{a^2 - 1} }} \cdot \frac{\sqrt{ a - \sqrt{a^2 - 1} }}{ \sqrt{ a - \sqrt{a^2 - 1} } }$

$\frac{1}{y} = \frac{ \sqrt{ a - \sqrt{a^2 - 1} }}{ \sqrt{a^2 - (a^2 - 1)}} = \sqrt{ a - \sqrt{a^2 - 1} }$

Putting this into the original equation gives:
$y^x + y^{-x} = 2 a$

Multiplying both sides by $y^x$ gives:

$y^{2x} + 1 = 2a y^x$

or
$y^{2x} - 2a y^x + 1 = 0$

This is a quadratic equation in $y^x$ so we may apply the quadratic formula:

$y^x = \frac{(2a) \pm \sqrt{(2a)^2 - (4)(1)(1)}}{(2 \cdot 1)}$

After some simplification we get
$y^x = a \pm \sqrt{a^2 - 1}$

So we have two cases:
$y^x = a + \sqrt{a^2 - 1} = y^2$, which leads to x = 2,

or
$y^x = a - \sqrt{a^2 - 1} = y^{-2}$, which leads to x = -2.

You can easily back-substitute and you will find that both solutions fit the original equation, so $x = \pm 2$.

-Dan

4. ## Re: Problem #19: Exponent Solution

But the answer should be $\pm 2$

$y^x+y^{-x}=y^2+y^{-2}$

$y^x-y^2+y^{-x}-y^{-2}=0$

$\left(y^x-y^2\right)\left(1-y^{-x}y^{-2}\right)=0$

which leads to $x=\pm 2$

5. ## Re: Problem #19: Exponent Solution

Originally Posted by Idea
But the answer should be $\pm 2$

$y^x+y^{-x}=y^2+y^{-2}$

$y^x-y^2+y^{-x}-y^{-2}=0$

$\left(y^x-y^2\right)\left(1-y^{-x}y^{-2}\right)=0$

which leads to $x=\pm 2$
I have fixed the problem in my derivation. Thanks for the catch!

-Dan