# Equsum Partitions of Sets of Positive Integers +
## 5 Aliquot Sets  
For any \(n\in\mathbb{Z}^{+}\), the _aliquot set_\(D^{\prime}(n):=\{d\in\mathbb{Z}^{+}:d\}\)_n_, \(d<n\}\) is the set of divisors less than \(n\), often called its _aliquot parts_. Its sum is \(s(n):=\Sigma D^{\prime}(n)=\sigma(n)-n\). Unlike \(\sigma(n)\), the function \(s(n)\) is not multiplicative, so it is less straightforward to use the structure of \(n\) to predict when \(s(n)\) will be a multiple of any given \(b\in\mathbb{Z}^{+}\). However, for the simplest case \(b=2,\ n\geq 3\), evidently \(2\big{|}s(n)\) if and only if \(\sigma(n)\) and \(n\) have the same parity: this occurs precisely when \(n\) is an odd square, or \(n\) is even and is neither a square nor twice a square. The construction used for Lemma 5 and Theorem 7 easily adapts to the aliquot case, and shows in particular that the aliquot set of an even perfect number has an equisum 2-partition. **Theorem 8**. : _Let \(a,\ m\in\mathbb{Z}^{+},\ 2\
mid m,\ p\in\mathbb{P},\ p\geq 3,\ 2^{a+1}\geq\sigma(p^{m})\). Then, the aliquot set \(D^{\prime}(2^{n}p^{m})\) has an equisum \(2\)-partition._  
Proof. : Let \(\mathcal{B}^{\prime}:=\left\{B_{1}^{\prime},B_{2}^{\prime}\right\}\) be the equisum 2-partition of the divisor set \(D(2^{n}p^{m})\) used to establish Theorem 7. Then \(\left\{2^{a-1}p^{i}:1\leq i\leq m\right\}\subset B_{1}^{\prime}\) and \(2^{n}p^{m}\in B_{2}^{\prime}\). Let  
\[B_{1}^{\prime\prime}:=B_{1}^{\prime}\backslash\left\{2^{a-1}p^{m}\right\},\ \ B_{2}^{\prime\prime}:=\left(B_{2}^{\prime}\backslash\left\{2^{a}p^{m}\right\} \right)\cup\left\{2^{a-1}p^{m}\right\}.\]  
Evidently, \(\mathcal{B}^{\prime\prime}:=\left\{B_{1}^{\prime\prime},B_{2}^{\prime\prime}\right\}\) is an equisum 2-partition of \(D^{\prime}(2^{n}p^{m})\). For any \(n\in\mathbb{Z}^{+}\), suppose the divisor set \(D(2n)\) has an equisum 2-partition \(\mathcal{B}:=\left\{B_{1},B_{2}\right\}\) such that \(n\in B_{1},2n\in B_{2}\). Then, the construction used for Theorem 8 modifies to yield an equisum 2-partition \(\mathcal{B}^{\prime}:=\left\{B_{1}^{\prime},B_{2}^{\prime}\right\}\) for the aliquot set \(D^{\prime}(2n)\), thus:  
\[B_{1}^{\prime}:=B_{1}\backslash\{n\},\ \ B_{2}^{\prime}:=\left(B_{2}\backslash \left\{2n\right\}\right)\cup\{n\}.\]  
Note that if \(\mathcal{B}\) is an equisum 2-partition of \(D(2n)\), the blocks of \(\mathcal{B}\) do not necessarily separate \(n\) and \(2n\). For instance, \(D(600)\) has an equisum 2-partition into [30, 300, 600] and its complement. On the other hand, the equisum 2-partition into [30, 100, 200, 600] and its complement does separate 300 and 600, so easily modifies to give an equisum 2-partition of \(D^{\prime}(600)\). For any \(a,b\in\mathbb{Z}^{+},\ p\in\mathbb{P},b\geq 2\), we have \(D^{\prime}(p^{a})=\{1\}\cup pD^{\prime}\left(p^{a-1}\right)\), so any \(b\)-partition of \(D^{\prime}(p^{a})\) has one block sum in the residue class \(1(\bmod p)\) and all others in \(0(\bmod p)\); hence, the block sums cannot be equal. Thus, if \(D^{\prime}(n)\) has an equisum \(b\)-partition, \(n\) must have a rank of at least 2. For brevity, we shall confine the remaining discussion to equisum 2-partitions of \(D^{\prime}(n)\) for odd \(n\) of rank 2. **Theorem 9**. : _Let \(n\in\mathbb{Z}^{+},\ 2\
mid n,\ \text{have just two distinct prime factors $p<q$}\). If the aliquot set \(D^{\prime}\!\left(n^{2}\right)\) has an equisum \(2\)-partition then \(p,\ q\) are twin primes or \(p=3,\ q=7\)._  
**Proof. ** Let \(n^{2}:=p^{2a}q^{2c}\) for some \(a,\ c\in\mathbb{Z}^{+}\). Clearly, \(D^{\prime}\big{(}n^{2}\big{)}\) has no equisum 2-partition if its largest aliquot divisor \(p^{2a-1}q^{2c}\) is greater than the sum of all other aliquot divisors. The sum of those other divisors is  
\[S:=\sigma\Big{(}p^{2a}q^{2c-1}\Big{)}+q^{2c}\sigma\Big{(}p^{2a-2}\Big{)}<\frac{ p^{2a+1}q^{2c}}{(p-1)(q-1)}+\frac{p^{2a-1}q^{2c}}{p-1}.\]  
Hence, \(S<p^{2a-1}q^{2c}\) certainly holds if \(p^{2}\leq(p-2)(q-1)\). For \(p\in\{3,\ 5\}\), this holds when \(q\geq 11\).