I am looking for conditions (if any are needed beyond properties of primes) for which $(\sum p_i)(\sum \frac{1}{p_i})$ over an arbitrary $i$ for a set of primes $\{p_i\}$ is unique in that there is no other set of primes $p_j$ for which $$(\sum p_i)(\sum \frac{1}{p_i}) = (\sum p_j)(\sum \frac{1}{p_j})$$
It is a bit difficult to describe so I will give an example:
I would like to know if there is a proof (if it is even true) that given any primes $p_1$ and $p_2$ that there are no other set of primes $\{p_3$,$p_4,..,p_k\}$ such that $$(p_1+p_2)(\frac{1}{p_1}+\frac{1}{p_2}) = (p_3+p_4+..+p_k)(\frac{1}{p_3}+\frac{1}{p_4}+..+\frac{1}{p_k})$$
And in general (again I don't know if this is true), that
$$(p_1+p_2+..+p_n)(\frac{1}{p_1}+\frac{1}{p_2}+..+\frac{1}{p_n}) \neq (p_3+p_4+..+p_s)(\frac{1}{p_3}+\frac{1}{p_4}+..+\frac{1}{p_s})$$
So basically whether or not I can find a set of two sets of primes of any length which satisfy the equality.
I have been trying to figure out a clever way to reduce this and would guess that there is some trick by the uniqueness of primes, but I don't see it.
Thanks,
Brian
$(p_1 + p_2) (1/p_1 + 1/p_2) = 2 + p_1/p_2 + p_2/p_1$ has $p$-adic valuation $p^1$ for $p = p_1$ and $p_2$, and not for any other prime. If this is the same as $(p_3 + \ldots + p_k)(1/p_3 + \ldots + 1/p_k)$, then certainly $p_1$ and $p_2$ must be in $\{p_3, \ldots, p_k\}$. So $(p_3 + \ldots + p_k)(1/p_3 + \ldots + 1/p_k) > (p_1 + p_2)(1/p_1 + 1/p_2)$, and they are not the same.
It's trickier for $n > 2$, because $p_1 + \ldots + p_n$ might be divisible by $p_1$ say.