Science:Math Exam Resources/Courses/MATH312/December 2009/Question 05 (a)
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Question 05 (a) 

Let denote the Mobius function. If is an integer, show that

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you? 
If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint. 
Hint 1 

Recall that the Mobius function is the function that evaluates to 1 if n is squarefree and has an even number of prime factors, 1 if n is squarefree and has an odd number of factors, and 0 if n has a square factor. First show/recall that the Mobius function is multiplicative. 
Hint 2 

Let . Show that this function is multiplicative then evaluate it on prime powers. Note: One could solve this problem by a more direct approach however learning about the theory of multiplicative functions is quite important and worth doing. 
Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work.

Solution 

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Please rate my easiness! It's quick and helps everyone guide their studies. First we show that the Mobius function is multiplicative. Let m and n be coprime integers. Then let
and . We wish to show that . Clearly, if m or n contains a square prime factor, then both sides evaluate to 0 so without loss of generality, suppose that each . We break into cases based on the number of prime factors. Case 1: M and N are both odd or both even.. In these cases, . Now, since the number of prime factors are both odd or both even, we have that and since these are now either plus or minus 1 (we've eliminated the case when the Mobius function is 0 on either side), we have
Case 2: M and N are of opposite parity.. Suppose that M is even (the argument is symmetric). In this case since we have an odd number of factors. Now, we have that and . Thus,
Hence the Mobius function is multiplicative. Now, let . I claim that this function is also multiplicative. For this, again let m and n be coprime integers (notice that ). We see that
Now as m and n are coprime, we see that for any divisor , we can write uniquely of the form where and and vice versa. Thus, we have
showing that this function is also multiplicative. So, by multiplicativity, it suffices to show that where p is a prime and a is an integer. Here, we have . Notice that the last sum might be nonexistent if a is 1. In any case, the last sum evaluates to 0 since all the terms contain a factor. Notice that and so so long as a is not 0 (which is allowed since n was an integer greater than 1). Thus we have a multiplicative function which evaluates to 0 on all prime powers, hence it must be zero for all values greater than 1 since . This completes the proof. 