How to show that if $\phi(n)$ equals to n itself, then n must be 1?

Question

That is:

If $\phi(n) = n$ then $n = 1$

Could someone give me a clue?

Answer 1

Hint:

Using the prime decomposition of $\;n\;$ we have:

$$n=\prod_{i=1}^k p_i^{a_i}\;,\;\;p_i\;\text{primes}\;,\;a_i\in\Bbb N\implies \phi(n)=n\prod_{i=1}^k\left(1-\frac1p\right)$$

Well, what those primes $\;p_i\;$ above could possibly be when $\;\phi(n)=n\;?\ldots$

Answer 2

Recall that $\varphi(n)$ is the number of integers $n$ in the interval $[0,n-1]$ that are relatively prime to $n$. If $n\gt 1$, then $0$ and $n$ are not relatively prime, so $\varphi(n)\le n-1$.