$(|x| + \sqrt {x ^ 2 - 1}) ^ x + (|x| - \sqrt {x ^ 2 - 1}) ^ x = 2(2x^2 - 1)$

Question

I have come across the equation $$(|x| + \sqrt {x ^ 2 - 1}) ^ x + (|x| - \sqrt {x ^ 2 - 1}) ^ x = 2(2x^2 - 1)$$. Of course we have $x \in (-\infty, -1) \cup(1, \infty)$. I noticed that $(|x| - \sqrt {x ^ 2 - 1}) ^ x = \frac{1} {(|x| + \sqrt {x ^ 2 - 1}) ^ x}$, but, still, I cannot find a way to write it more easier. I tried proving the fact that there are 2 functions with different monotonies, but that is not true.

Answer 1

Consider $A:=|x|+\sqrt{x^2-1}$; then we have, $$A^{-1}=\frac{1}{|x|+\sqrt{x^2-1}}=\frac{|x|-\sqrt{x^2-1}}{x^2-(x^2-1)}=|x|-\sqrt{x^2-1}$$

So, your equation is essentially,

$$A^x+A^{-x}=2(2x^2-1)$$

and note that we have $A+A^{-1}=2|x|$ and $A-A^{-1}=2\sqrt{x^2-1}$

Can you take it from here?

Answer 2

Hint:

Let $|x|=a\ge0, \sqrt{x^2-1}=b\implies a^2-b^2=1, a^2+b^2=?$

If $x\ge0, x=a$

$$(a+b)^a+(a-b)^a=2(a^2+b^2)=(a+b)^2+(a-b)^2$$

$$\implies (a+b)^a+\dfrac1{(a+b)^a}=(a+b)^2+\dfrac1{(a+b)^2}$$

If $p+\dfrac1p=q+\dfrac1q, p=q$ or $p=\dfrac1q$

What if $x<0?$

Answer 3

Hint:

Notice that with exponent $2$,

$$(|x| + \sqrt {x ^ 2 - 1}) ^ 2 + (|x| - \sqrt {x ^ 2 - 1}) ^ 2= 2(2x^2 - 1),$$

so $x=2$ is a solution.

And as

$$(|x| + \sqrt {x ^ 2 - 1}) (|x| - \sqrt {x ^ 2 - 1})=1,$$ the function is even, so $-2$ is another solution.

With exponent $1$,

$$2|x|=2(2x^2-1)$$

and $|x|=1$ is also a solution.

Remains to show if they are the only ones. From the first identity, it is obvious that for $x>2$,

$$(|x| + \sqrt {x ^ 2 - 1}) ^ x + (|x| - \sqrt {x ^ 2 - 1}) ^ x\\>(|x| + \sqrt {x ^ 2 - 1}) ^ 2 + (|x| - \sqrt {x ^ 2 - 1}) ^ 2= 2(2x^2 - 1).$$