Science:Math Exam Resources/Courses/MATH312/December 2012/Question 04

MATH312 December 2012

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Question 04
Define the sum of divisors function $\displaystyle \sigma (n)$ and number of divisors function $\displaystyle \tau (n)$. Show that there is no positive integer n with $\displaystyle \phi (n)=14.$

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

For the last part, use the fact that $\displaystyle \phi (n)=n\prod _{p\mid n}\left(1-{\frac {1}{p}}\right)=\prod _{p^{\alpha }\parallel n}p^{\alpha -1}(p-1)=14$. where $\displaystyle n=\prod _{p^{\alpha }\parallel n}p^{\alpha }$ and where we recall that $\displaystyle p^{\alpha }\parallel n$ means that $\displaystyle p^{\alpha }$ fully divides n (that is, this is the highest power of p that divides n).

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. First, $\displaystyle \sigma (n)=\sum _{d\mid n}d$ and $\displaystyle \tau (n)=\sum _{d\mid n}1$ gives the required definitions. For the last part, we proceed by a contradiction and suppose that $\displaystyle \phi (n)=14$ for some integer n. As stated in the hint, we have $\displaystyle \phi (n)=n\prod _{p\mid n}\left(1-{\frac {1}{p}}\right)=\prod _{p^{\alpha }\parallel n}p^{\alpha _{p}-1}(p-1)=14$. where $\displaystyle n=\prod p^{\alpha _{p}}$. Now every odd prime present in the factorization of n contributes a power of two to $\displaystyle \phi (n)$ due to the $\displaystyle (p-1)$ factor. Thus, there can be at most one odd prime dividing n, say $\displaystyle p$. If $\displaystyle p$ occurs to a power of two, then because it divides the left hand side, it must divide 14 and so must be 7. The prime $\displaystyle p$ cannot occur to a third power or higher since otherwise there are two factors of $\displaystyle p$ on the left and at most one on the right. However $\displaystyle \phi (7^{2})=42>14$ and so the power must be 1. Thus $\displaystyle n=2^{\alpha }p$ for some $\displaystyle \alpha$. Notice that as the phi-function is multiplicative, we have that $\displaystyle 14=\phi (n)=\phi (2^{\alpha })\phi (p)=2^{\alpha -1}(p-1)$ (or we have that $\displaystyle \phi (n)=p-1$ if no power of 2 divides n) Thus, as $\displaystyle p-1$ is even, we have that $\displaystyle \alpha =1$ (or n is odd) and further that $\displaystyle p=15$ which is not a prime. This is a contradiction and hence this equation has no solution.

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