Mathematical Reflections 2013, Issue 1 - Problem S255

Problem:
Solve in real numbers the equation $$2^x+2^{-x}+3^x+3^{-x}+\left(\dfrac{2}{3}\right)^x+\left(\dfrac{2}{3}\right)^{-x}=9x^4-7x^2+6.$$

Proposed by Mihaly Bencze.

Solution:
We rewrite the equation in the form $$\left(2^{x/2}-2^{-x/2}\right)^2+\left(3^{x/2}-3^{-x/2}\right)^2+\left((2/3)^{x/2}-(2/3)^{-x/2}\right)^2=9x^4-7x^2.$$ Let $$f(x)=\left(2^{x/2}-2^{-x/2}\right)^2+\left(3^{x/2}-3^{-x/2}\right)^2+\left((2/3)^{x/2}-(2/3)^{-x/2}\right)^2$$ and $$g(x)=9x^4-7x^2.$$ Since $f(x)$ and $g(x)$ are even functions, it suffices to find the solutions when $x \geq 0$. It is easy to see that $x=0$ and $x=1$ are solutions of the given equation. Moreover, $f(x)$ is increasing for $x \geq 0$ since it is a sum of increasing functions, and $g(x)$ is increasing if $x \geq \sqrt{7/18}$ and decreasing if $0 \leq x \leq \sqrt{7/18}$, as can be seen from $g'(x)$. Since $f(x)$ and $g(x)$ are both injective functions if $x \geq 1$, then $h(x)=g(x)-f(x)$ is an injective function if $x \geq 1$, so $h(1)=0$ and $h(x) \neq 0$ for all $x>1$. Therefore the equation has no other solutions for $x \geq 0$, which means that the only solutions of the equation are $x=-1,0,1$.