- From NRICH,

find all real solutions to

- $ \left(2-x^2\right)^{x^2-3 \sqrt{2} x+4}=1\, $

Extension: What if x is permitted to be a complex number?

- MathHelp replies,

### First tryEdit

*a*^{b}=1 when *a*=1 or *b*=0 (and *a* is nonzero), so the first try at a solution is to solve

- $ x^2-3 \sqrt{2} x+4=0\, $

and

- $ 2-x^2=1\, $

and take the union of the results.
The first equation gives us $ \sqrt{2} $ and $ \sqrt{8} $. The second equation gives us 1 and −1. But then we need to throw out $ \sqrt{2} $ because that solution gives us 0^{0}, which is undefined. So the results of the first try are:

- $ <math>\sqrt{8}, \ 1, \ \mbox{and} \ 1. \, $

### Second tryEdit

Noting that (-1)^{2}=1 and that $ e^{2 i \pi k}=1\, $ for all integers, *k*, there are certainly other solutions to *a*^{b}=1. For all I know at this point, there may be real solutions in x to the pair of equations

- $ x^2-3 \sqrt{2} x+4=a\, $

and

- $ 2-x^2=b,\, $

where *a* and *b* are complex solutions to *a ^{b}*=1.

So I would like to start by characterizing solutions to *a ^{b}*=1. First, the equation is solved whenever

*a*=1, regardless of

*b*. If

*a*is not equal to 1, then,

*a*= e^{b}^{b ln(a)}= $ \mathrm{e}^{b \ln{a} - 2 i \pi k}=1\, $

and so the complete solution to *a ^{b}*=1 is

- $ a=1,\, $
- $ a \ne 0, \ a \ne 1, \ b = \frac{2 i \pi k}{\ln(a)}\, $

The second solution can also be expressed as

- $ a = e^{\frac{2 i \pi k}{b} }\, $

^{[improvement needed: You can help]}