# 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?

• 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 1 at 11:09
• @Batominovski thank you, I've got it now. – MartinYakuza Sep 1 at 11:15

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.

• The $\sin, \cos$ relation is a fantastic spot. – Teresa Lisbon Sep 1 at 15:39

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!

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.