Science:Math Exam Resources/Courses/MATH437/December 2011/Question 02
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Question 02 

Let , with . If given by is a multiplicative function, show that and that for . 
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 hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! 
Hint 

Argue using the fact that
where . Your proof should also have a feel of analysis in it. 
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Solution 

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Please rate my easiness! It's quick and helps everyone guide their studies. First we show that . Let be an integer. Since our function is multiplicative, we know that . Expanding gives
Isolating for gives
Dividing both sides by gives
Making sufficiently large, we can see that the right hand side tends to 0. Since the left hand side is an integer, we see that the right hand side must in fact be 0 and hence so must the left hand side. Thus,
giving since as stated in the question. Next, we show that all the other coefficients must be 0. Let j be the largest index less that k such that . We assume that j exists and find a contradiction. Thus, . Let m and n be arbitrary nonzero, nonunital positive integers such that . Then, we see again that
and hence
Cancelling and isolating for yields
As before dividing by and letting tend to infinity shows that the modified right hand side is 0 and thus, we have
Hence, either or . However, is completely independent of . Thus as j is distinct from k (so ), we must have that which shows that no such j can exists. Hence each and every for must in fact be 0. This completes the proof. Keep in mind that and this holds with this polynomial. Also it is clear that is multiplicative. 