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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

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