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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?

Viera Čerňanová
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MartinYakuza
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    Rewrite the equation in the form $a^x+b^x=1$, where $0<a<b<1$. Note that the function $f:\mathbb{R}\to\mathbb{R}$ defined by $f(t):=a^t+b^t\text{ for all }t\in\mathbb{R}$ is strictly decreasing with $f(2)=1$. – Batominovski Sep 01 '20 at 11:09
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    @Batominovski thank you, I've got it now. – MartinYakuza Sep 01 '20 at 11:15

3 Answers3

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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.

Marek Kryspin
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3

Another way.

Rewrite our equation in the following form: $$\left(\sqrt{\frac{2-\sqrt2}{2+\sqrt2}}\right)^x+1=\left(\frac{2}{\sqrt{2+\sqrt{2}}}\right)^x.$$ We see the the left side decreases and the right side increases,

which says that our equation has one real root maximum.

But $2$ is a root and we are done!

Michael Rozenberg
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Divide your original equation by $2^x$ getting

$((\sqrt{2+\sqrt2}/2)^x+((\sqrt{2-\sqrt2}/2)^x=1.$

As $\sqrt2<2$, both terms on the left side are exponential functions with base strictly between $0$ and $1$. Therefore, the left side is strictly decreasing forcing no more than one solution. Thus any solution that might be found by inspection must be the only one.

Oscar Lanzi
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