20.18*. (New York, 77). Find all differentiable functions $f: \mathbf{R} \longrightarrow \mathbf{R}$, satisfying the identity

$$
f^{\prime}\left(\frac{x+y}{2}\right)=\frac{f(y)-f(x)}{y-x}, \quad x, y \in \mathbf{R}, \quad x \neq y
$$
20.18. Since, by the condition of the problem, for any value of $y \neq 0$ the identity

$$
f^{\prime}(x)=\frac{f(x+y)-f(x-y)}{2 y} \quad(x \in \mathbf{R})
$$

holds, and the right-hand side is differentiable with respect to $x$, then

$$
\begin{aligned}
& f^{\prime \prime}(x)=\frac{f^{\prime}(x+y)-f^{\prime}(x-y)}{2 y} \Rightarrow \\
& \quad=\frac{1}{2 y}\left[\frac{f(x+2 y)-f(x)}{2 y}-\frac{f(x-2 y)-f(x)}{(-2 y)}\right] \Rightarrow \\
& =\frac{f(x+2 y)+f(x-2 y)-2 f(x)}{4 y^{2}}
\end{aligned}
$$

Further, the last expression is again differentiable with respect to $x$, so $f^{\prime \prime \prime}(x)=\frac{f^{\prime}(x+2 y)+f^{\prime}(x-2 y)-2 f^{\prime}(x)}{4 y^{2}}=$ $=\frac{1}{4 y^{2}}\left[\frac{f(x+4 y)-f(x)}{4 y}+\frac{f(x-4 y)-f(x)}{(-4 y)}-\frac{f(x+4 y)-f(x-4 y)}{4 y}\right] \equiv 0$. Thus, the sought functions must satisfy the identity $f^{\prime \prime \prime}(x) \equiv 0$, therefore

$$
\begin{gathered}
f^{\prime \prime}(x) \equiv F^{\prime \prime}(0), \quad f^{\prime}(x) \equiv f^{\prime \prime}(0) x+f^{\prime}(0) \\
f(x)=f^{\prime \prime}(0) x^{2} / 2+f^{\prime}(0) x+f(0)
\end{gathered}
$$

Note that any function of the form

$$
f(x)=a x^{2}+b x+c \text {, where } a, b, c \in \mathbf{R},
$$

possesses all the properties indicated in the problem.