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(D) Let $f(x) = ax + b$. Given the equation $f(f(x)) = x + f(x)$. Substituting $f(x)$ into the equation: $a(ax + b) + b = x + (ax + b)$ $a^2x + ab + b

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$2$

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(D) Let $f(x) = ax + b$. Given the equation $f(f(x)) = x + f(x)$. Substituting $f(x)$ into the equation: $a(ax + b) + b = x + (ax + b)$ $a^2x + ab + b = x + ax + b$ $a^2x + ab = x + ax$ $x(a^2 - a - 1) + ab = 0$. For this to hold for all $x$, the coefficients must be zero: $a^2 - a - 1 = 0$ and $ab = 0$. Since $a^2 - a - 1 = 0$, $a$ cannot be $0$, therefore $b = 0$. Solving $a^2 - a - 1 = 0$ using the quadratic formula: $a = \frac{1 \pm \sqrt{5}}{2}$. Thus, the functions are $f(x) = \left(\frac{1 + \sqrt{5}}{2}\right)x$ and $f(x) = \left(\frac{1 - \sqrt{5}}{2}\right)x$. There are $2$ such real linear functions.

The number of real linear functions $f(x)$ satisfying $f(f(x))=x+f(x)$ is

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