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Find all the numbers $n$ such that $\varphi(n)=5$

Attempt:

$$n=\prod\limits_{i=1}^{k} p_i^{\beta_i}$$

$$\varphi(n)=\prod\limits_{i=1}^{k}p^{\beta_i-1}(p-1)$$

We need:

$$\prod\limits_{i=1}^{k}p^{\beta_i-1}(p-1)=5$$

Since$$\varphi(n)\leqslant n-1$$

there isn't such $n$ is this correct?

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    There are only two positive integers $n$ such that $\phi(n)$ is odd. – Batominovski Jul 03 '16 at 19:49
  • Not instantly clear to me why you point out that $\varphi(n)≤n-1$. From the prior line we see that $p$ can't be odd (lest $p-1$ be even) so $n$ would have to be a power of $2$, but it is easy to rule that case out as well. – lulu Jul 03 '16 at 19:49
  • $$ \varphi(n) \geq \sqrt {\frac{n}{2}} $$ and many other LOWER bounds. http://math.stackexchange.com/questions/301837/is-the-euler-phi-function-bounded-below/301856#301856 – Will Jagy Jul 03 '16 at 20:15
  • $$ \varphi(n) \geq 8 \cdot \left( \frac{n}{30} \right)^{7/8} $$ says that we do not need to check numbers larger than $29.$ Looking a bit harder, we can stop at $17.$ – Will Jagy Jul 03 '16 at 20:26

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Suppose that $n$ has an odd prime divisor. Then $$ \phi(n)=n\prod_{p\mid n}\frac{p-1}{p}=\frac{n}{\prod_{p\mid n}p}\cdot \prod_{p\mid n}(p-1). $$ Both factors are integers, and the second factor is even, because $p-1$ is even for our odd prime divisor $p$. Hence $\phi(n)$ is even, so different from $5$. Now we consider the remaining case, that $n$ has no odd prime divisor. Then $n=2^r$. For $r=1$ we have $\phi(n)=\phi(2)=1$, and for $r\ge 2$ we have $\phi(n)=2^r-2^{r-1}$, which is even, so different from $5$.

Summarizing we can say that there is no $n$ with $\phi(n)=5$ (or any other odd number greater than $1$, for that matter).

Dietrich Burde
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