Let $\phi(n) $ be Euler's Totient Function

Let us consider $$ |\{ n \in \mathbb{N} : \phi (n) = 8 \} | = 5, $$ and $$ |\{ n \in \mathbb{N} : \phi (n) = 14 \} | = 0. $$

How would I go about proving this?


If there are $r\ge1$ odd primes that divide $n$, then $2^r$ divides $\phi(n)$ (*). And if $2^t$ ($t\ge1$) divides $n$ then $2^{t-1}$ divides $\phi(n)$.

Therefore, if $\phi(n)=14$, then $n$ is the power of an odd prime ($n=p^s$) or its double ($n=2p^s$). Any case, $\phi(n)=p^{s-1}(p-1)$ so we have two possibilities:

  • If $s>1$ then $p^{s-1}$ divides $14$, that is, $p=7$ and $s=2$, but $\phi(7^2)=42$.
  • If $s=1$ then $p=n$ and $\phi(n)=n-1$, but $15$ is not prime.

Thus, there is no $n$ such that $\phi(n)=14$.

Can you now try with $8$? It's not very different.

Proof of (*): Let $p_1,\ldots,p_r$ be the odd prime factors of $n$. Then $$n=2^k\prod_{j=1}^r p_j^{\alpha_j}$$ for some $k\ge 0$ and $$\phi(n)=\phi(2^k)\prod_{j=1}^r(p_j-1)p_j^{\alpha_j-1}$$

Sinnce every $p_j-1$ is even, $2^r$ divides $\phi(n)$.

  • $\begingroup$ "If there are $r\ge 1$ odd primes that divide n" is it the true for every n > 2? $\endgroup$ – user180834 Jan 30 '15 at 20:25
  • $\begingroup$ There is no solution to $\phi(n)=78$, is it? $\endgroup$ – ms._VerkhovtsevaKatya Jan 29 at 11:07

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