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!