5. Let $A$ be a set of $m$ real numbers, with the sum of its elements being $a$; $B$ be a set of $n$ real numbers, with the sum of its elements being $b$. Define
$$
C=\{\boldsymbol{x} \mid \boldsymbol{x}=(s, t), s \in A, t \in B\} \text {. }
$$

Let $\boldsymbol{x} y$ each take all vectors in $C$ (allowing $\boldsymbol{x}=\boldsymbol{y}$) and compute $\boldsymbol{x} \cdot \boldsymbol{y}$. Then the sum of these dot products $\sum_{x, y \in C} \boldsymbol{x} \cdot \boldsymbol{y}$ equals $\qquad$ (the result should be expressed in terms of $a$, $b$, $m$, and $n$).
5. $n^{2} a^{2}+m^{2} b^{2}$.

From the coordinate form of vectors, we get
$$
\begin{array}{l}
\sum_{\boldsymbol{x} \in C} \boldsymbol{x}=\sum_{s \in A} \sum_{t \in B}(s, t)=\sum_{s \in A}\left(\sum_{t \in B} s, \sum_{t \in B} t\right) \\
=\sum_{s \in A}(n s, b)=\left(n \sum_{s \in A} s, \sum_{s \in A} b\right) \\
=(n a, m b) . \\
\text { Therefore, } \sum_{x=y \in C} \boldsymbol{x} \cdot \boldsymbol{y}=\left(\sum_{x \in C} \boldsymbol{x}\right) \cdot\left(\sum_{y \in C} \boldsymbol{y}\right) \\
=(n a, m b) \cdot(n a, m b)=n^{2} a^{2}+m^{2} b^{2} .
\end{array}
$$