Science:Math Exam Resources/Courses/MATH437/December 2011/Question 02
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Let , with . If given by
is a multiplicative function, show that and that for .
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Argue using the fact that
where . Your proof should also have a feel of analysis in it.
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First we show that . Let be an integer. Since our function is multiplicative, we know that
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
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.