# Find all positive integers $n$ such that $\phi(n)=10$ [duplicate]

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Find all positive integers n such that $\phi(n)=10$.

I have found 2 positive integers: $11$ and $22$.

Is there any more?

Thanks!

## marked as duplicate by Dietrich Burde, kingW3, GNUSupporter 8964民主女神 地下教會, Leucippus, NamasteDec 20 '17 at 0:52

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• I know what you mean by ϕ, but some people here don't. It'd be best to define it in the question – Wen Dec 19 '17 at 11:29
• – lab bhattacharjee Dec 19 '17 at 11:30

## 2 Answers

Write $n=p_1^{a_1}p_2^{a_2}\dots p_i^{a_i}$, where $p_1,p_2,\dots,p_i$ are distinct primes. $\phi(n)=\phi(p_1^{a_1}p_2^{a_2}\dots p_i^{a_i})=(p_1^{a_1}-p_1^{a_1-1})(p_2^{a_2}-p_2^{a_2-1})\dots(p_i^{a_i}-p_i^{a_i-1})=10$

if, say, $p_1^{a_1}-p_1^{a_1-1}=1$, then $p_1=2$ and $a_1=1$. If $p_1^{a_1}-p_1^{a_1-1}=10$, what can you say about $p_1$ and $a_1$?

Now, $10=2\times5$ is the only nontrivial factorization of $10$.

So in this case either $n=p_1^{a_1}p_2^{a_2}p_3^{a_3}$ where $p_1^{a_1}-p_1^{a_1-1}=1, p_2^{a_2}-p_2^{a_2-1}=2, p_3^{a_3}-p_3^{a_3-1}=5$, or $n=p_1^{a_1}p_2^{a_2}$ where $p_1^{a_1}-p_1^{a_1-1}=2, p_2^{a_2}-p_2^{a_2-1}=5$. There is no such $n$ which satisfies either of these.

Conclude that $n=11,22$ are the only solutions.

Hint: Let $p$ be a prime.

• If $p$ divides $n$, then $p-1$ divides $\phi(n)$.

• If $p^2$ divides $n$, then $p$ divides $\phi(n)$.

You can also use that $\phi(n) \geq \frac{\sqrt{n}}{\sqrt{2}}$ (see this question) and so limit a brute force search to $n\le 200$.