[proofplan] We prove both identities by counting the same finite set of additive quadruples in two different ways. First, we partition the quadruples according to their common sum $x=a_1+b_1=a_2+b_2$, giving the second moment of $r_{A+B}$. Then we rewrite the same equation as $a_1-a_2=b_2-b_1$ and partition by the common difference $t$, giving the product of the two difference representation functions. [/proofplan] [step:Partition additive quadruples by their common sum] Let \begin{align*} \mathcal{Q} := \{(a_1,a_2,b_1,b_2) \in A \times A \times B \times B : a_1 + b_1 = a_2 + b_2\}. \end{align*} By definition, $E^+(A,B)=\#\mathcal{Q}$. For each $x \in G$, define \begin{align*} \mathcal{Q}_x := \{(a_1,a_2,b_1,b_2) \in A \times A \times B \times B : a_1+b_1=x \text{ and } a_2+b_2=x\}. \end{align*} The sets $\mathcal{Q}_x$ are pairwise disjoint as $x$ varies over $G$, and their union is $\mathcal{Q}$, because every quadruple in $\mathcal{Q}$ has a unique common sum $x=a_1+b_1=a_2+b_2$. Hence \begin{align*} E^+(A,B) = \#\mathcal{Q} = \sum_{x \in G} \#\mathcal{Q}_x. \end{align*} For fixed $x \in G$, choosing an element of $\mathcal{Q}_x$ is the same as choosing two ordered representations of $x$ as a sum $a+b$ with $a \in A$ and $b \in B$. Therefore \begin{align*} \#\mathcal{Q}_x = r_{A+B}(x)^2. \end{align*} Substituting this into the preceding count gives \begin{align*} E^+(A,B)=\sum_{x \in G} r_{A+B}(x)^2. \end{align*} The sum is finite because $r_{A+B}(x)=0$ unless $x \in A+B$, and $A+B$ is finite. [/step] [step:Rewrite the additive equation as an equality of differences] For any quadruple $(a_1,a_2,b_1,b_2) \in A \times A \times B \times B$, the equation \begin{align*} a_1+b_1=a_2+b_2 \end{align*} is equivalent, by adding $-a_2-b_1$ to both sides in the abelian group $G$, to \begin{align*} a_1-a_2=b_2-b_1. \end{align*} Thus \begin{align*} \mathcal{Q} = \{(a_1,a_2,b_1,b_2) \in A \times A \times B \times B : a_1-a_2=b_2-b_1\}. \end{align*} [/step] [step:Partition additive quadruples by their common difference] For each $t \in G$, define \begin{align*} \mathcal{D}_t := \{(a_1,a_2,b_1,b_2) \in A \times A \times B \times B : a_1-a_2=t \text{ and } b_2-b_1=t\}. \end{align*} The sets $\mathcal{D}_t$ are pairwise disjoint as $t$ varies over $G$, and their union is $\mathcal{Q}$ by the difference formulation from the previous step. Therefore \begin{align*} E^+(A,B) = \#\mathcal{Q} = \sum_{t \in G} \#\mathcal{D}_t. \end{align*} Fix $t \in G$. The number of ordered pairs $(a_1,a_2) \in A \times A$ satisfying $a_1-a_2=t$ is $r_{A-A}(t)$. The number of ordered pairs $(b_1,b_2) \in B \times B$ satisfying $b_2-b_1=t$ is also $r_{B-B}(t)$, because the map \begin{align*} \{(b_1,b_2) \in B \times B : b_2-b_1=t\} &\to \{(u,v) \in B \times B : u-v=t\} \\ (b_1,b_2)&\mapsto (b_2,b_1) \end{align*} is a bijection. Hence \begin{align*} \#\mathcal{D}_t = r_{A-A}(t)r_{B-B}(t). \end{align*} Substituting into the partition count yields \begin{align*} E^+(A,B)=\sum_{t \in G} r_{A-A}(t)r_{B-B}(t). \end{align*} The sum is finite because $r_{A-A}(t)=0$ unless $t \in A-A$, and $A-A$ is finite. This proves both identities. [/step]