Yesterday, I saw the following Riemann zeta function identity:
I took some time to try to derive it myself and to my great pleasure, I succeeded.
Eventually, I realized that it suffices to show that
are equal as multisets. As sets, they are both representations of the set of -tuples of positive integers such that the third is a multiple of the least common multiple of the first two. In the latter one, the frequency of is the number of that divides both and such that . In the other one, if we write as where , the condition equates to , which corresponds to the number of dividing and and such that and with that, both dividing , which is the frequency of via the former representation.
The coefficients of the Dirichlet series of the LHS of that identity can be decomposed as follows:
The coefficients of the Dirichlet series of the RHS of that identity are
Observe how both are equivalent in that via the multiset equivalence proved above, determines the same multiset of for both and across that, the values of the same function are summed. Hence the two series are equal.
I read it a couple days ago and actually remembered it this time in a way that I will never forget it. It invokes Euclid’s lemma, which states that if for prime, then or , which can be proved using Bezout’s lemma. For existence, it does induction on the number of factors, with as the trivial base case. For the non base case, wherein our number is composite, apply the inductive hypothesis on the factors. For uniqueness, assume two distinct factorizations: . By Euclid’s lemma, each of the s divides and is thus equal to one of the s. Keep invoking Euclid’s lemma, canceling out a prime factor on each iteration and eventually we must end with in order for the two sides to be equal.