In analogy to perfect numbers, let's say I wanted a set $S = \{s_1, s_2,..., s_n\}$ of numbers with the property:
$$\frac{s_1+s_2+\cdots s_n}{n}= \sigma(S) := \sum_{d\mid s_1\wedge d\mid s_2, d\mid s_3,\cdots \wedge d\mid s_n}_{d\ne s_1, d\ne s_2, \cdots d\ne s_n}d$$
that is, the average of $S$ equals the sum of the proper divisors common to all the elements of $S$.
- Can such a "perfect set" even exist?
There is the obvious case of just repeatedly including the same perfect number as many times as you like (for example: $\{28, 28, 28, 28, ...\}$). Then this reduces to the usual definition of perfect number.
I suspect, but have no idea how to show, that this is actually the only type of example. Heuristically, it seems like the set of divisors common to everything would have to be "bunched together" in a very specific way, in order for $\sigma(S)$ to be large enough to reach the average of $S$, and this is probably just too delicate. But my intuition could be wrong.
- Alternately: what happens if we take the geometric mean instead of the average? Since this can be so much smaller.
The product of all $n$ numbers would then have to be a perfect $n$th power for this to make sense, which seems like a pretty strict requirement, too. But, assuming this, does such a set then become possible?
More generally, each $s \in S$ comes with a set of divisors. If we collect all of these sets into (for lack of a better term) a "divisor system," what can be said about the combinatorics of such a system? The above sum $\sigma(S)$ concerns the sum of the elements in the intersection of these sets; what about their symmetric difference? etc. Is there a field of number theory which deals with these questions?
You can simplify your first problem using Bezout's identity to finding solutions $\{ s_1,s_2,s_3\cdots s_n \}$ such that:
$$\gcd(s_1,s_2,\cdots s_n)\sum_{s_i= \gcd(s_1,s_2,s_3,\cdots s_n)}_{1\leq i \leq n}1+\frac{s_1+s_2+\cdots s_n}{n}=\sigma(\gcd(s_1,s_2,\cdots s_n))$$
This is because:
$$\sum_{d\mid s_1\wedge d\mid s_2, d\mid s_3,\cdots \wedge d\mid s_n}_{d\ne s_1, d\ne s_2, \cdots d\ne s_n}d=\sigma(\gcd(s_1,s_2,\cdots s_n))-\sum_{s_i\mid \gcd(s_1,s_2,s_3,\cdots s_n)}_{1\leq i \leq n}s_i=$$
$$\sigma(\gcd(s_1,s_2,\cdots s_n))-\sum_{s_i= \gcd(s_1,s_2,s_3,\cdots s_n)}_{1\leq i \leq n}s_i=$$
$$\sigma(\gcd(s_1,s_2,\cdots s_n))-\gcd(s_1,s_2,\cdots s_n)\sum_{s_i= \gcd(s_1,s_2,s_3,\cdots s_n)}_{1\leq i \leq n}1$$
Sense if $s_i\mid \gcd(s_1,s_2,s_3,\cdots s_n)$ that means we have $s_i\leq \gcd(s_1,s_2,s_3,\cdots s_n)$ but we know that $\gcd(s_1,s_2,s_3,\cdots s_n)\ge s_i$ sense $\gcd(s_1,s_2,s_3,\cdots s_n)\mid s_i$.
Thus such an $s_i$ must be equal to $\gcd(s_1,s_2,s_3,\cdots s_n)$.