How can I show that $\phi(n)=14$ is impossible? Here, $\phi$ is Euler's phi-function.

Question

hero · Answer

pre-algebra, I'm never helping you with any of your questions.

anonymous · Answer

Why not? :(

hero · Answer

Because, everytime you post a question, it is something I've never seen before.

hero · Answer

What course are you taking?

anonymous · Answer

I have an idea for this one:

Suppose that there is a number $n$ such that $\phi(n)=14$. Then $n$ has prime-power factorization $n=p_{1}^{a_{1}}p_{2}^{a_{2}}\cdots p_{k}^{a_{k}}$ and $\phi(n)=\phi(p_{1}^{a_{1}}p_{2}^{a_{2}}\cdots p_{k}^{a_{k}})=\phi(p_{1}^{a_{1}})\phi(p_{2}^{a_{2}})\cdots \phi(p_{k}^{a_{k}})=14$. Moreover, the prime-power factorization of $14$ is $14=2\cdot7$. We then need (by definition) two numbers of the form $p^a(p-1)$ with $p$ prime and $a$ a positive integer. Clearly, $2=2^1(2-1)$, but 7=p^a(p-1) would imply that $7$ is composite since $2$ is the smallest prime such that $2-1=1$ and $2^3=8>7$.

This is elementary number theory.

hero · Answer

That would explain why I could never help you.