# Computing the Möbius function

1. Sep 17, 2008

### CRGreathouse

From the Mathematica documentation:
SquareFreeQ[n] checks to see if n has a square prime factor. This is done by computing MoebiusMu[n] and seeing if the result is zero; if it is, then n is not squarefree, otherwise it is. Computing MoebiusMu[n] involves finding the smallest prime factor q of n. If n has a small prime factor (less than or equal to 1223), this is very fast. Otherwise, FactorInteger is used to find q.

How does this work? I can't think of a way to compute mu(n) faster than:
* finding a small p for which p^2 | n, or
* finding that n is prime, or
* checking that for all p < (n)^(1/3), p^2 does not divide n
* factoring n, which can be faster than the third option for large n

Edit: I checked my copy of "Open problems in number theoretic complexity, II", which has O8: Is SQUAREFREES in P? Assuming that this is still open, any algorithm known would either be superpolynomial or of unproven complexity. But even an exponential-time improvement on the naive algorithm would be of interest.

Last edited: Sep 17, 2008
2. Sep 22, 2008

### CRGreathouse

The interesting line, to me, is this:
If n has a small prime factor (less than or equal to 1223), this is very fast.

This seems a fairly strong claim. In particular, it would mean that the "very fast" algorithm could distinguish between n = 2pq and n = 2pq^2 for large p and q.

I saw a reference to this at http://www.marmet.org/louis/sqfgap/ :

3. Sep 22, 2008

### Hurkyl

Staff Emeritus
The implication cannot be true, because it would mean that if N doesn't have a small factor, that we could compute $\mu(N) = -\mu(3N)$, where the latter can be done rapidly.

4. Sep 23, 2008

### CRGreathouse

Yeah, I figured as much, thanks.

So factoring n is pretty much the fastest way to find mu(n), right? Except in the special cases I outlined above?

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