5. 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
$$
\begin{array}{l}
(f(x)+f(z))(f(y)+f(t)) \\
=f(x y-z t)+f(x t+y z) .
\end{array}
$$
5. Suppose for all $x, y, z, t \in \mathbf{R}$,
$$
\begin{array}{l}
f(x y-z t)+f(x t+y z) \\
=(f(x)+f(z))(f(y)+f(t)) .
\end{array}
$$

Let $x=y=z=0$. Then
$$
2 f(0)=2 f(0)(f(0)+f(t)) \text {. }
$$

In particular, $2 f(0)=4 f^{2}(0)$, so,
$$
f(0)=0 \text { or } f(0)=\frac{1}{2} \text {. }
$$

If $f(0)=\frac{1}{2}$, then $f(0)+f(t)=1$, which gives
$$
f(x)=\frac{1}{2} \text {. }
$$

If $f(0)=0$, let $z=t=0$ in (1), then
$$
f(x y)=f(x) f(y) .
$$

In particular, $f(1)=f^{2}(1)$, so,
$$
f(1)=0 \text { or } f(1)=1 \text {. }
$$

If $f(1)=0$, then $f(x)=f(x) f(1)=0, x \in \mathbf{R}$.

Thus, assume $f(0)=0, f(1)=1$. Let $x=0, y=t=1$,
then (1) gives
$$
f(-z)+f(z)=2 f(z),
$$

i.e., for $z \in \mathbf{R}, f(-z)=f(z), f(x)$ is an even function.
Since $f\left(x^{2}\right)=f^{2}(x) \geqslant 0$ and $f(x)$ is an even function, then for $y \in \mathbf{R}$, we have
$$
f(y) \geqslant 0 .
$$

Let $t=x, z=y$ in (1), then
$$
f\left(x^{2}+y^{2}\right)=(f(x)+f(y))^{2} \text {, }
$$

i.e., $f\left(x^{2}+y^{2}\right) \geqslant f^{2}(x)=f\left(x^{2}\right)$.
Therefore, for $u \geqslant v \geqslant 0, f(u) \geqslant f(v)$, and for $x \in \mathbf{R}^{+}, f(x)$ is an increasing function.
Let $y=z=t=1$ in (1), then
$$
f(x-1)+f(x+1)=2(f(x)+1) .
$$

We can prove by induction that $f(n)=n^{2}$ for non-negative integers $n$.
Since $f(x)$ is an even function, then $f(n)=n^{2}$ for all integers. By $f(x y)=f(x) f(y)$, we get $f(a)=a^{2}$ for all rational numbers.
Assume for some positive $x, f(x) \neq x^{2}$.
If $f(x)a>\sqrt{f(x)}$, then $f(a)=a^{2}>f(x)$.
By the increasing property of $f(x)$, we have $f(a) \leqslant f(x)$, a contradiction.
For $f(x)>x^{2}$, a similar proof can be given.
Thus, $f(x)=x^{2}, x \in \mathbf{R}^{+}$.
Since $f(x)$ is an even function, we have $f(x)=x^{2}, x \in \mathbf{R}$.
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 conditions. Hence, they are the solutions.