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Can anyone help me with this!? If $n=p_1^{k_1},p_2^{k_2},\ldots $ Then I applied the given condition of divisibility of $\varphi(n)$ but can't reach to a conclusion.

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4 Answers 4

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Hint: If $p^2 \mid n$, then $p \mid \phi(n)$.

Indeed, if $p^e$ is the power of $p$ appearing in the factorization of $n$, then $\phi(n)=\phi(p^e)\cdots=p^{e-1}(p-1)\cdots$.

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  • $\begingroup$ Can you explain how? $\endgroup$
    – Planck
    Commented Feb 4, 2019 at 9:55
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First, note that $\phi$ multiplicative. Then if $n=p_1^{k_1}\cdots p_m^{k_m}$, then $\phi(n)=\phi(p_1^{k_1})\cdots\phi(p_m^{k_m})$.

I believe you can prove that if for some prime $p$ and some $k\in\mathbb{Z^+}$, and $k>1$, then $p|\phi(p^k)$.

So suppose that some $k_i$ is greater than $1$. Then $p_i|\phi(p_i^{k_i})\hspace{0.2cm}|\phi(n)\hspace{0.2cm}|n-1$. But also, $p|n$. So, prime $p$ should be equal to $1$. Absurd....

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If $p^2 \mid n$, then $p \mid \varphi(n)$, but obviously $p \nmid n-1$ and so $\varphi(n) \nmid n-1$.

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we know that $n-\varphi(n)=\Pi\, p_i^{k_i-1}(\Pi\,p_i-\Pi\,(p_i-1))$ so if there is $k_i>1$ on the one hand we have $p_i|n-\varphi(n)$ and in the second one we know that $p_i|n$ then, $p_i|\varphi(n)$. So $p_i|n-1$ and $p_i|n$ ---> absurd so $k_i=1~~\forall i$

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