1. **Problem Statement:** Find the number of real-valued solutions to the equation $$2^x + 2^{-x} = 2 - (x-2)^2.$$\n\n2. **Rewrite the equation:** Let us analyze the left side first. Recall that $$2^{-x} = \frac{1}{2^x}.$$ So the left side is $$2^x + \frac{1}{2^x}.$$\n\n3. **Use substitution:** Let $$y = 2^x > 0.$$ Then the equation becomes $$y + \frac{1}{y} = 2 - (x-2)^2.$$\n\n4. **Analyze the left side:** The expression $$y + \frac{1}{y}$$ for $$y > 0$$ satisfies the inequality $$y + \frac{1}{y} \geq 2$$ by AM-GM inequality, with equality only when $$y=1$$ (i.e., $$x=0$$).\n\n5. **Analyze the right side:** The right side is $$2 - (x-2)^2.$$ Since $$(x-2)^2 \geq 0,$$ the maximum value of the right side is 2, attained at $$x=2.$$\n\n6. **Compare both sides:** The left side $$\geq 2$$ and the right side $$\leq 2.$$ For equality, both sides must equal 2.\n\n7. **Check equality conditions:**\n- Left side equals 2 when $$y=1 \Rightarrow 2^x=1 \Rightarrow x=0.$$\n- Right side equals 2 when $$(x-2)^2=0 \Rightarrow x=2.$$\n\nSince these values of $$x$$ are different, the equation cannot hold with equality.\n\n8. **Check if solutions exist:**\n- For $$x=0,$$ left side is 2, right side is $$2 - (0-2)^2 = 2 - 4 = -2,$$ no equality.\n- For $$x=2,$$ right side is 2, left side is $$2^2 + 2^{-2} = 4 + \frac{1}{4} = 4.25,$$ no equality.\n\n9. **Check behavior for other values:**\n- Left side $$\geq 2$$ always.\n- Right side $$\leq 2$$ always.\n- The right side decreases as $$|x-2|$$ increases, becoming negative for large $$|x|.$$\n\n10. **Graphical intuition:** The left side is always at least 2, the right side is at most 2 and decreases quadratically. The two curves do not intersect.\n\n**Conclusion:** There are no real solutions to the equation.\n\n**Final answer:** 0