Online Course Discussion Forum
MCIII 4.5 and 4.6 and 4.7
Hello!
I have no idea how to do 4.5. I tried plugging in functions but it just got very tedious.
For 4.6 I realized that 3 cannot divide (pi-1) for any i. But I'm not sure how to find the prime numbers, I know they have to be congruent to 2 mod 3, but I listed 2, 5, 11, 17, 23, 29, 41, 47, 53, 39, 71, and there seems to be infinitely many more. How should I proceed?
For 4.7 I tried analyzing both sides as the number of numbers that are relatively prime to something else, and I seem to be getting a contradiction and I don't know how to solve this problem.
Thank you!
Tina Jin
4.5: It does help to evaluate for $n=1, 2, 3, 4$ and see if there is a pattern. To make it less tedious, you can evaluate $\mu(n)$ first, then the Mobius transform of each $n$ is the sum of certain $\mu(k)$ values. After trying small values, you may want to try $p^k$, powers of primes.
4.6: We are not limited to prime numbers. Your conclusion is correct that for any prime factor $p_i$ of $n$, $3$ cannot divide $p_i - 1$. So you can say the number $n$ does not have prime factors of the form $3k+1$ for any $k$. However, this is not all the restrictions. If you examine the formula for $\phi(n)$ carefully, you need that $n$ cannot have two factors of $3$. Thus you also add the requirement that $9\nmid n$.
4.7: Use the formula $$\phi(n) = n\prod_{i=1}^k \left( 1 - \frac{1}{p_i}\right),$$ where $p_i$ are the prime factors of $n$.
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