Let f(x)=x^2-2 for real x. The number of real values of x for which f(f(f(x)))=x but f(x)≠ x is ______.

Any real $x$ with $f^{3}(x)=x$ must satisfy $|x|\le2$: if $|x|>2$ then $f(x)>2$, and $f$ is strictly increasing and larger than the argument on $(2,\infty)$, so the iterates run away and never return. So put $x=2\cos\theta$ with $\theta\in[0,\pi]$. Then $f(x)=4\cos^2\theta-2=2\cos2\theta$, and $f^{3}(x)=2\cos8\theta$. The equation $f^3(x)=x$ becomes $\cos8\theta=\cos\theta$, giving $8\theta=\pm\theta+2\pi m$, i.e. $\theta=\dfrac{2\pi m}{7}$ or $\theta=\dfrac{2\pi m}{9}$ in $[0,\pi]$. First family: $m=0,1,2,3$ gives $4$ values; second: $m=0,1,2,3,4$ gives $5$ values; $\theta=0$ is common, so there are $8$ distinct solutions (matching degree $8$). Fixed points of $f$ itself: $x^2-x-2=0$, $x=2,-1$, which are two of these $8$. Excluding them leaves $8-2=6$ points of exact period $3$.