# For any positive integer n, show that µ(n)*µ(n+1)*µ(n+2)*µ(n+3)=0

*print*Print*list*Cite

### 1 Answer

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`

**Sources:**