The equation $a^x +a^{-x}=b$ Given $a$ is a fixed real number that is greater than one, how many real numbers $b$ are there such that the equation $^{x} + a^{-x} = $  has a unique real solution x?
To solve this problem, I tried to plug in $a = 3$.
So, I want to find how many reals $b$ there are such that $3^{x} + 3^{-x} = b$ has a unique real solution x.
Setting $3^x$ = $t$, I got that $t + \frac{1}{t} = b$. It follows that $t^2 - tb + 1 = 0$. From this, it can be obtained that $t = \frac{b \pm \sqrt{b^2-4}}{2}$.
I set the discriminant equal to zero: $b^{2}-4=0$, so $b = \pm 2$. This means that $t = \pm 1$. Since $t = 3^{x}$ (previously defined), $x$ is real only when $t = 1$.
From the work, doesn't it follow that there is only one real number $b$ (that is when $b$ = 2) such that there is one unique real solution $x$? But the answer I see says there are an infinite values of $b$ such that there is one real solution x.
What am I missing in my analysis?

If we let $y=a^x$, this becomes
$$y+\frac1y=b,~~y>0$$
$$y^2-by+1=0$$
$$y=\frac{b\pm\sqrt{b^2-4}}2$$
For there to be one solution, $b^2-4=0$, so $b=\pm2$.  Clearly, $b>0$, since $y>0$, so the only option is $b=2$.

Alternatively, one could note that for every $x$ that satisfies the equality, $-x$ satisfies the equality, hence one must have $x=0$, and thus $b=2$.
