# Prove that $\phi (p-1) \leq (p-1)/2$, where $p\geq 3$ is a prime

Prove that $$\phi (p-1) \leq (p-1)/2$$, where $$p\geq 3$$ is a prime

My idea:

We know that $$\phi (p^k) = p^{k-1} (p-1)$$. Then to use the $$\phi$$ is multiplicative. now I have $$\phi (p-1)$$ in my hand, but then how to complete? could anyone help me, please?

Edit:
OP didn't check for $$p=2$$, the only prime where above $$\phi (p-1) \leq (p-1)/2$$ holds to fail.

Note that for every $$p\geq 3$$, $$(p-1)$$ is an even number.
Therefore, by fundamental Theorem of Arithmetic, it will have a unique factorization which contains $$2^i$$, $$i\geq 1$$.

Now, Note that $$\phi(2^i)=2^{i-1}\text{ or }\phi(2^i)=\frac{2^i}{2}$$.

Therefore, using multipicative nature of $$\phi$$, we have $$\phi(p-1)=\phi(2^i)\phi(\frac{p-1}{2^i}) \leq \frac{p-1}{2}$$ Case $$1$$ : $$(p-1)$$ is an exact power of $$2$$ i.e. $$\frac{p-1}{2^i}=1$$

$$\phi(p-1)=\phi(2^i)\phi(1) = \frac{p-1}{2}$$ Case $$2$$ : $$\frac{p-1}{2^i}=p_{1}^{\alpha_{1}}p_{2}^{\alpha_{2}}...p_{n}^{\alpha_{n}}$$, where $$p_{j}\geq 3$$ $$\forall j=1,2,...,n$$ and $$\alpha_{j}\geq 0$$ $$\forall j=1,2,...,n$$ and alteast one of $$\alpha_{j}\geq 1$$. $$\phi(\frac{p-1}{2^i})=\phi(p_{1}^{\alpha_{1}}p_{2}^{\alpha_{2}}...p_{n}^{\alpha_{n}})=\phi(p_{1}^{\alpha_{1}})\phi(p_{2}^{\alpha_{2}})...\phi(p_{n}^{\alpha_{n}}) < \frac{p-1}{2^i}$$

Now, Can you complete Case $$2$$ and combine it with Case $$1$$ to get the desired answer?

Note: I have used the convention, $$\phi(1)=1$$.

Hint:

Two even numbers can't be coprime.

• Then what is next? Jun 25, 2019 at 1:46

Hint: If $$p$$ is odd, then $$p-1$$ is even. If $$p$$ is even, then $$p=2$$ and $$\phi(1)$$ is defined to be $$1$$ by convention and so exceeds $$1/2$$.

• Thank you for clarifying this case Jun 25, 2019 at 2:05