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For a positive integer n:
`mu(n)` = 1 if n has an even number of prime factors and none of them is a square.
`mu(n) = -1` if n has an odd number of prime factors and none of them are squares.
`mu(n) = 0` if n has factors that are squares.
Now for any number n, one of the four consecutive numbers n, (n+1), (n+2) or (n+3) will be divisible by 4. As 4 is a square number and `mu(n) = 0` if n has a factor that is a square one of the values `mu(n), mu(n+1),mu(n+2) or mu(n+3)` is equal to zero.
This proves that for any number n `mu(n)*mu(n+1)*mu(n+2)*mu(n+3) = 0`
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