<s>[INST] Find all $x\in\mathbb{R}$ such that $\left( \sqrt{2-\sqrt{2} }\right)^x+\left( \sqrt{2+\sqrt{2} }\right)^x=2^x$. Find all $x\in\mathbb{R}$ such that:
$$
\left( \sqrt{2-\sqrt{2} }\right)^x+\left( \sqrt{2+\sqrt{2} }\right)^x=2^x\,.
$$

Immediately we notice that $x=2$ satisfies the equation.
Then we see that $LHS=a^x+b^x$, where $a<1$ and $b<2$, therefore $RHS$ grows faster (for larger $x$, $LHS\approx b^x<2^x$)
Hence $x=2$ is the only real solution.

Unfortunately I don't know whether this line of reasoning is correct. Moreover, if it is indeed correct, how to write this formally?
 [/INST] It is not difficult (formula for the double angle) to show that $$\sin \left(\frac{ \pi }{8} \right)= \sqrt{ \frac{2- \sqrt{2} }{4} }$$ which in combination with the trigonometric one gives $$\cos^2\left(\frac{ \pi }{8} \right)=1-\sin^2\left(\frac{ \pi }{8} \right)=\frac{2+ \sqrt{2} }{4}\Rightarrow \cos \left(  \frac{ \pi }{8} \right)= \sqrt{  \frac{2+ \sqrt{2} }{4} }$$ thus our equation can be expressed equivalently in the form $$\left( \sqrt{2-\sqrt{2} }\right)^x+\left( \sqrt{2+\sqrt{2} }\right)^x=2^x$$ $$\left( \sqrt{\frac{2- \sqrt{2} }{4}}\right)^x+\left( \sqrt{\frac{2+ \sqrt{2} }{4}}\right)^x=1 $$ $$\sin^x\left(  \frac{ \pi }{8} \right)+\cos^x \left(  \frac{ \pi }{8} \right)=1$$ of course thanks to the trigonometric one $x=2$ is a trivial solution. Uniqueness of this solution is due to the fact $\sin \& \cos \le 1$. Formally, you can consider cases $x>2$ or $x<2$ and estimate the left side.
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