Inequality $(\sqrt5+2)^x+(\sqrt5-2)^x\le2\sqrt5$

Question

What is the number of all integer solutions to the inequality: $$(\sqrt5+2)^x+(\sqrt5-2)^x\le2\sqrt5$$

I'm pretty clueless to what to do with this one.

This is what I tried:

$(\sqrt5+2)^x+(\sqrt5-2)^x\le2\sqrt5\\(\sqrt5+2)^x\le2\sqrt5-(\sqrt5-2)^x$

then I do $\sqrt[x]{}$ over both sides, to get

$\sqrt5+2\le\sqrt[x]{2\sqrt5-(\sqrt5-2)^x}$

but I'm pretty sure I can't do anything with this.

Second idea I had:

$t=\sqrt5+2\\t^x+(t-4)^x\le2\sqrt5$

but again I think I can't do anything this way.

Answer 1

Note that $$(\sqrt{5}-2)(\sqrt{5}+2)=1$$ So you will get $$(\sqrt{5}-2)^x=\frac{1}{(\sqrt{5}+2)^x}$$

Answer 2

Let $f(x)$ be the LHS of the inequality.

Since $(\sqrt 5+2)(\sqrt 5-2)=1$, we see that $$f(-x)=f(x)$$

Also since we have $$(\sqrt 5+2)^2=9+4\sqrt 5\gt 2\sqrt 5$$ we have $$f(x)\gt 2\sqrt 5$$ for $|x|\ge 2$.

Hence, we have to have $x=\pm 1,0$ and these are sufficient.