# If $\varphi(n)\mid n-1$ then prove that there exist no prime such that $p^2\mid n$

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.

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 $$\phi(n)=\phi(p^e)\cdots=p^{e-1}(p-1)\cdots$$.

• Can you explain how? – Planck Feb 4 at 9:55

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....

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

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 $$k_i=1~~\forall i$$