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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

ab=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 00, 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 ab=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 ab=1.

So I would like to start by characterizing solutions to ab=1. First, the equation is solved whenever a=1, regardless of b. If a is not equal to 1, then,

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

and so the complete solution to ab=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]

External referencesEdit