F19 (43-5, India) Find all functions $f$ from the set of real numbers $\mathbf{R}$ to $\mathbf{R}$ such that for all $x, y, z, t \in \mathbf{R}$, we have
$$
(f(x)+f(z))(f(y)+f(t))=f(x y-z t)+f(x t+y z) .
$$
Let the given equation be denoted as (*). Substituting \( x = y = z = 0 \) into (*), we get
$$
2 f(0)(f(0) + f(t)) = 2 f(0),
$$

Substituting \( t = 0 \) into the above equation, we get \( 4(f(0))^2 = 2 f(0) \), so
$$
f(0) = 0 \text{ or } f(0) = \frac{1}{2}.
$$

If \( f(0) = \frac{1}{2} \), substituting into (1), we get
$$
f(0) + f(t) = 1, f(t) = \frac{1}{2},
$$

i.e., \( f(x) = \frac{1}{2} \).
If \( f(0) = 0 \), substituting \( z = t = 0 \) into (*), we get
$$
f(x) f(y) = f(xy),
$$

Substituting \( x = y = 1 \), we get \( (f(1))^2 = f(1) \), so \( f(1) = 0 \) or \( f(1) = 1 \).
When \( f(1) = 0 \), from (2) we get
$$
f(x) = f(x) f(1) = 0, x \in \mathbf{R}.
$$

When \( f(1) = 1 \), substituting \( x = 0, y = t = 1 \) into (*), we get
$$
\begin{aligned}
2 f(z) & = f(-z) + f(z), \\
f(z) & = f(-z), z \in \mathbf{R}.
\end{aligned}
$$

Thus, \( f(x) \) is an even function.
Since substituting \( x = y, z = t = 0 \) into (*), we get
$$
f(x^2) = (f(x))^2 \geqslant 0,
$$

and \( f(x) \) is an even function, so for all \( y \in \mathbf{R} \), \( f(y) \geqslant 0 \).
Substituting \( x = t, y = z \) into (*), we get
$$
(f(x) + f(y))^2 = f(x^2 + y^2),
$$

so
$$
\begin{aligned}
f(x^2 + y^2) & = (f(x))^2 + 2 f(x) f(y) + (f(y))^2 \\
& \geqslant (f(x))^2 \\
& = f(x^2),
\end{aligned}
$$

Thus, when \( u \geqslant v \geqslant 0 \), \( f(u) \geqslant f(v) \), i.e., for \( x \in \mathbf{R}^+, f(x) \) is increasing.
Substituting \( y = z = t = 1 \) into (*), we get
$$
2(f(x) + 1) = f(x-1) + f(x+1),
$$

Using (3) and mathematical induction, it is easy to show that for non-negative integers \( n, f(n) = n^2 \) holds. From \( f(xy) = f(x) f(y) \), we get \( f(a) = a^2 \) for all positive rational numbers \( a \). Next, we prove that \( f(x) = x^2, x \in \mathbf{R}^+ \). If not, there exists some \( x_0 \in \mathbf{R} \) such that \( f(x_0) \neq x_0^2 \).

If \( f(x_0) < x_0^2 \), by the increasing property of \( f \), we have \( a \geqslant x_0 \), which is a contradiction.
If \( f(x_0) > x_0^2 \), similarly, we get a contradiction.
Therefore, \( f(x) = x^2, x \in \mathbf{R}^+ \). Since \( f(x) \) is an even function, \( f(x) = x^2, x \in \mathbf{R} \).

Upon verification, \( f(x) = 0, x \in \mathbf{R}; f(x) = \frac{1}{2}, x \in \mathbf{R}; f(x) = x^2, x \in \mathbf{R} \) all satisfy the given conditions, so they are the solutions.