Numbers obtainable in two ways as a sum over complementary subsets of a set of primes Let $\{p_1,p_2,\ldots ,p_r\}$ be a set of distinct primes.
Denote the set $\{1,2,\ldots ,r\}$ by $C$.
Now, let $A,B\subset C$ such that
$$A\cup B=C, \quad A\cap B=\emptyset.$$
These sets are called $\textit{complement}$.
Now let 
$$\sum_{i\in A,j\in B}(p_i-1)(p_j-1)=N.$$
My question is, are there other other complement subsets of $D,E\subset C$ ($A\neq D,E$) such that 
$$\sum_{i\in D,j\in E}(p_i-1)(p_j-1)=N.$$
If such sets exist what is the minimal order of $C$ which admits such a phenomenon?
Thanks in advance for any help.

It is possible to see that 
$$\sum_{i\in A, j\in B} (p_i-1)(p_j-1) = ((\sum_{i\in A} p_i) - \#A)((\sum_{i\in B} p_i) - \#B).$$ 
This means that if we find two sets of primes adding up to some $n$, we can construct such an example. One can take $7+13=20$ and $3+17=20$. 
Take then $C=\{ 3,5,7,13,17 \}$ and $A = \{3,17\}$ and $B=\{5,7,13\}$, we obtain $(20-2)(25-3)$.
But the sets $A'=\{3,5,17 \}$ and $B'=\{7,13\}$ render the same number.
So indeed this can be done.
