Discussion
If \(\mu\) is the Mobius function and \(n\) is a positive integer, then \(\mu(n)\ \mu(n + 1)\ \mu(n + 2)\ \mu(n + 3)\) is equal to :
Answer: B
The product is 0.
Among any four consecutive integers \(n, n + 1, n + 2, n + 3\), one is divisible by 4. That number has the square factor \(2^{2}\), so the Mobius function is 0 there, and the whole product is 0.
The Mobius function only takes the values 0, 1 and -1, so a product of four values can never be 4 or 6. A value of 1 would need all four numbers to be square-free, which the divisibility by 4 rules out.
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