Science:Math Exam Resources/Courses/MATH312/December 2010/Question 05

MATH312 December 2010

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Question 05
Find all positive integers n such that $\displaystyle \phi (n)=22$ and prove that there are no others. Here $\displaystyle \phi$ denotes the Euler $\displaystyle \phi$ function.

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
Recall that $\displaystyle \phi (n)=n\prod _{p\mid n}\left(1-{\frac {1}{p}}\right)$ Start with this and use divisibility rules to help.

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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. We begin by noting that $\displaystyle 2\cdot 11=22=\phi (n)=n\prod _{p\mid n}\left(1-{\frac {1}{p}}\right)=n\prod _{p\mid n}{\frac {p-1}{p}}=\prod _{p_{i}^{\alpha _{i}}\parallel n}p_{i}^{\alpha _{i}-1}(p-1)$ where above we used the Fundamental Theorem of Arithmetic to write (recall that $\displaystyle \parallel$ means fully divides) $\displaystyle n=\prod _{p_{i}^{\alpha _{i}}\parallel n}p_{i}^{\alpha _{i}}.$ Our main equation of discussion will be $\displaystyle 2\cdot 11=\prod _{p_{i}^{\alpha _{i}}\parallel n}p_{i}^{\alpha _{i}-1}(p_{i}-1)$. We proceed in cases. First we discuss the powers of 2 in n. Case 1: $\displaystyle 8\mid n$. In this case, notice that 4 divides the right hand side of the main equation but only 2 divides the left hand side. This is a contradiction. Case 2: $\displaystyle 4\parallel n$. In this case, notice that 2 divides the right hand side of the main equation and 2 divides the left hand side. However, if another odd prime divides n, then the right hand side is divisible by an extra power of 2 which is a contradiction. The remaining subcase in this case is if no odd primes divde n, that is, $\displaystyle n=4$. But since also $\displaystyle \phi (4)=2\neq 22$ this case cannot occur. Case 3: $\displaystyle 2\parallel n$ or n is odd . In either of these cases, since $\displaystyle \phi$ is multiplicative and as $\displaystyle \phi (2)=1$, we have that either of these cases can occur, that is, if $\displaystyle \phi (n)=22$, then $\displaystyle \phi (2n)=22.$ Therefore, without loss of generality, we suppose 2 does not divide $\displaystyle n$. Now we deal with the odd primes. Notice that for each odd prime, we get a factor of 2 on the right hand side of the main equation. Thus only exactly one odd prime can occur. Thus $\displaystyle n=p^{\alpha }$. Since only 2 and 11 occur as prime factors on the left hand side of the equation, we know that we must have that $\displaystyle \alpha \leq 2$ (the right hand side of the main equation has an $\displaystyle p^{\alpha -1}$ term so the factor could be 2). If $\displaystyle \alpha =2$, then the prime p in question must be 11 since p divides the right hand side. So $\displaystyle n=11^{2}$ but this is inadmissable since $\displaystyle \phi (11^{2})\neq 22$. Thus n is a prime. For primes, we have that $\displaystyle \phi (p)=p-1$. This value we know is 22 and so $\displaystyle n=23$. Thus combining the two discussions shows us that either $\displaystyle n=23$ or $\displaystyle n=46$ completing the question.

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