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From this question I'm trying to check if the inverse sentence works:

If $\phi(d)|\phi(n)$ then $d|n$

So I know that

If $\phi(d)|\phi(n)$ then $\phi(n)=k\phi(d)$

Then

$$ \phi(d)=d(1-1/p_1)(1-1/p_2)...(1-1/p_r)$$ $$ k\phi(d)=k[d(1-1/p_1)(1-1/p_2)...(1-1/p_r)]\not \not = kd$$ Then $$d \not| n$$

If is correct please let me know, if not, any hint or help will be really appreciated.

mraz
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2 Answers2

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If $\phi(n) = k\phi(d)$, then it is not true that $n$ must be exactly $kd$ for the same $k$ : it could $ld$ for some other $l$, and it would still qualify as a multiple of $d$. So your working is incomplete.

Having said that, before going to the definitions itself, you should always try out your conjecture for small numbers : in this case, the quantities are easy to compute, and you can at least see if what you are saying is worth the while.

For example, in the above case I will tell you this much : I can find two single digit numbers $d,n$ such that $\phi(d)$ divides $\phi(n)$ but $d$ does not divide $n$. So the converse is clearly not true.

Now, can you figure out one such pair? For an even bigger hint :

$d =4$, now figure out some $n$ giving a counterexample.

Sarvesh Ravichandran Iyer
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If $n = \prod p_i^{k_i}$ is the prime factorization of $n$, then $\phi(n) = \prod (p_i -1)\prod p_i^{k_i - 1}$.

It's easy to is if $\prod p_i^{j_i}|\prod q_m^{k_m} \implies j_i \le k_i; \{p_i\}\subset \{q_i\} \implies \prod (p_i -1)\prod p_i^{j_i - 1}|\prod (q_m -1)\prod q_i^{m_i - 1}$.

But the converse need not apply, and it'd be hypothetically possible for some of prime factors of $p_i -1$ to be some of the primes $q_m$ that don't correspond to the $p_i$..

Example: $4= (5-1)5^0 = \phi(5)$ and $4 = (2-1)2^2 = \phi 8$. So $\phi(8)|\phi(5)$ but $8\not \mid 5$.

fleablood
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