Mobius Function Proof
Prove that there are infinitely many integers n such that:
μ(n) + μ(n+1) = 0?
***μ(n) represents the Mobius Function
I've been trying to use the Well-Ordering Axiom but to no success. I attempted to have a closed set and prove that there was a greater number that did not appear in the set but I'm having trouble.
Any help would be appreciated. Thanks!
Originally Posted by FatherMike
The equality is equivalent to proving that either there are infinitely many integers n s.t. both n and n+1 aren't square free, or else there are inf. many integers n s.t. both n and n+1 are square free and some of them has an odd number of prime divisors more than the other one, and I think we can do it as follows: take the sequence of odd squares:
. There are infinite such squares, and for each such square n, its antecesor n-1 is zero multiple 4 and thus divisible by a square and thus you have inf. many pairs of consecutive naturals at which the mu function vanishes!