# Equsum Partitions of Sets of Positive Integers +
## 5 Aliquot Sets
For \(p\geq 7\), it holds when \(q\geq p+4\); since \(q\) is prime, the only case not then excluded is \(q=p+2\). \(\Box\)  
**Theorem 10**. : _For every \(a,\ c\in\mathbb{Z}^{+}\), the aliquot set \(D^{\prime}\big{(}3^{2a}5^{2c}\big{)}\) has an equisum 2-partition._  
**Proof. ** When \(n=3^{2}5^{2}=225\), the aliquot set \(D^{\prime}(n)\) has the equisum 2-partition  
\[D^{\prime}(225)=\{1,\ 3,\ 15,\ 25,\ 45\}\cup\{5,\ 9,\ 75\}.\]  
Fix \(c\in\mathbb{Z}^{+}\) and suppose \(\mathcal{B}:=\{B_{1},B_{2}\}\) is an equisum 2-partition of \(D^{\prime}\big{(}3^{2}5^{2c}\big{)}\) with \(3:=5^{2c}\in B_{1}\). Since  
\[D^{\prime}\big{(}3^{2}5^{2c+2}\big{)}\backslash D^{\prime}\big{(}3^{2}5^{2c} \big{)}=\big{\{}3^{2}5^{2c}\big{\}}\cup 5^{2c+1}D^{\prime}\big{(}3^{2}5\big{)},\]  
then \(D^{\prime}\big{(}3^{2}5^{2c+2}\big{)}\) has 2-partition \(\mathcal{B}^{\prime}:=\big{\{}B^{\prime}_{1},B^{\prime}_{2}\big{\}}\) such that  
\[ \begin{array}{l}B^{\prime}_{1}:=\Big{(}B_{1}\backslash\big{\{}3\cdot 5^{2c} \big{)}\backslash\cup\big{\{}3\cdot 5^{2c+1},3\cdot 5^{2c+2}\big{\}},\\ B^{\prime}_{2}:=B_{2}\cup\big{\{}3\cdot 5^{2c},3^{2}5^{2c}\big{\}}\cup\big{\{}5^{2c +1},5^{2c+2},3^{2}5^{2c+1}\big{\}}.\end{array} \]  
Moreover, \(\mathcal{B}^{\prime}\) is equisum because  
\[\Sigma B^{\prime}_{1}-\Sigma B_{1}-3\cdot 5^{2c}=87\cdot 5^{2c},\ \ \Sigma B^{ \prime}_{2}-\Sigma B_{2}=87\cdot 5^{2c}.\]  
Now fix \(a,\ c\in\mathbb{Z}^{+}\) and suppose \(\mathcal{C}:=\{C_{1},C_{2}\}\) is an equisum 2-partition of \(D^{\prime}\big{(}3^{2a}5^{2c}\big{)}\) with \(3^{2a}\in C_{1}\). Then \(D^{\prime}\big{(}3^{2a+2}5^{2c}\big{)}\) has 2-partition \(\mathcal{C}^{\prime}:=\big{\{}C^{\prime}_{1},C^{\prime}_{2}\big{\}}\) such that  
\[ \begin{array}{l}C^{\prime}_{1}:=\Big{(}C_{1}\backslash\big{\{}3^{2a}\big{\}} \cup 3^{2a+1}E\Big{(}5^{2c}\big{)}\cup 3^{2a+2}E\Big{(}5^{2c-2}\Big{)},\\ C^{\prime}_{2}:=C_{2}\cup\Big{\{}3^{2a},3^{2a}5^{2c}\big{\}}\cup 3^{2a+1}5E \Big{(}5^{2c-2}\Big{)}\cup 3^{2a+2}5E\Big{(}5^{2c-2}\Big{)},\end{array} \]  
where \(E\Big{(}5^{2c}\big{)}:=\Big{\{}5^{2i}:i\in\mathbb{Z},\ 0\leq i\leq c\Big{\}}\). Furthermore, \(\mathcal{C}^{\prime}\) is equisum because  
\[\Sigma C^{\prime}_{1}-\Sigma C_{1}-3^{2a}=7K\cdot 3^{2a},\ \Sigma B^{\prime}_{2}- \Sigma B_{2}=7K\cdot 3^{2a},\ \ K:=\Big{(}5^{2c}-1\Big{)}/2.\]  
Induction on \(a\) and \(c\) completes the argument. \(\Box\)  
As before, the proof of Theorem 10, and that which follows for Theorem 11, do essentially provide unconditional algorithms for constructing an equisum partition in any concrete instance. **Theorem 11**. : _For \(a\in\{1,\ 2\}\) and every \(c\in\mathbb{Z}^{+}\), the aliquot set \(D^{\prime}\big{(}3^{2a}7^{2c}\big{)}\) has an equisum 2-partition._  
**Proof. ** When \(n=3^{2}7^{2}=441\), the aliquot set \(D^{\prime}(n)\) has the equisum 2-partition  
\[D^{\prime}(441)=\{3,\ 147\}\cup\{1,\ 7,\ 9,\ 21,\ 49,\ 63\}.\]Fix \(c\in\mathbb{Z}^{+}\) and suppose \(\mathcal{B}:=\{B_{1},B_{2}\}\) is an equisum 2-partition of \(D^{\prime}\big{(}3^{2}7^{2c}\big{)}\) with \(7^{2c}\in B_{2}\). Then \(D^{\prime}\big{(}3^{2}7^{2c+2}\big{)}\) has 2-partition \(\mathcal{B}^{\prime}:=\big{\{}B^{\prime}_{1},B^{\prime}_{2}\big{\}}\) such that  
\[ \begin{array}{c}B^{\prime}_{1}:=B_{1}\cup\big{\{}7^{2c},\ 3\cdot 7^{2c+2} \big{\}},\\ B^{\prime}_{2}:=\big{(}B_{2}\big{\backslash}\Big{\{}7^{2c}\big{\}}\big{)}\cup \big{\{}3^{2}7^{2c},\ 7^{2c+1},\ 3\cdot 7^{2c+1},\ 3^{2}7^{2c+1},\ 7^{2c+2}\big{\}}.\end{array} \]  
Moreover \(\mathcal{B}^{\prime}\) is equisum because  
\[\Sigma B^{\prime}_{1}-\Sigma B_{1}=148\cdot 7^{2c},\ \ \Sigma B^{\prime}_{2}- \Sigma B_{2}-7^{2c}=148\cdot 7^{2c}.\]  
When \(n=3^{4}7^{2}=3969\), the aliquot set \(D^{\prime}(n)\) has an equisum 2-partition comprising  
\[\{1,\ 7,\ 49,\ 63,\ 147,\ 189,\ 441,\ 567\}\]  
and its complement. When \(n=3^{4}7^{4}=194481\), the aliquot set \(D^{\prime}(n)\) has an equisum 2-partition comprising \(\left\{1,\ 3^{3}7,\ 3\cdot 7^{4},\ 3^{3}7^{4}\right\}\) and its complement.