# Totient function and divisibility

• Sep 27th 2013, 03:05 PM
topsquark
Totient function and divisibility
The problem:
Quote:

Prove that if d divides n then $\phi (d)$ divides $\phi (n)$.
(I will presume that d and n are positive integers.) This is one of those "duh!" thoughts but I am messing up the proof somehow. I'd appreciate it is someone could take a look at it.

First: If d|n then n = kd for some positive integer k. Using their decomposition into powers of primes, let
$d = 2^{a_2} \cdot 3^{a_3} \cdot 5^{a_5} \cdot \text{ ...}$
and
$k = 2^{b_2} \cdot 3^{b_3} \cdot 5^{b_5} \cdot \text{ ...}$

Then
$kd = 2^{a_2 + b_2} \cdot 3^{a_3 + b_3} \cdot 5^{a_5 + b_5} \cdot \text{ ...}$

Now, we have that
$\phi (d) = \left ( 2^{a_2 -1 } \right ) (2 - 1) \cdot \left ( 3^{a_3 -1 } \right ) (3 - 1) \cdot \left ( 5^{a_5 -1 } \right ) (5 - 1) \cdot \text{ ...}$

and
$\phi (kd) = \left ( 2^{a_2 + b_2 -1 } \right ) (2 - 1) \cdot \left ( 3^{a_3 + b_3 - 1 } \right ) (3 - 1) \cdot \left ( 5^{a_5 + b_5 - 1 } \right ) (5 - 1)\cdot \text{ ...}$

So finally we get
$\frac{ \phi (kd) }{ \phi (d) } = \left ( 2^{(a_2 + b_2 - 1) - (a_2 - 1)} \right ) \cdot \left ( 3^{(a_3 + b_3 - 1) - (a_3 - 1)} \right ) \cdot \text{ ...}$

$\frac{ \phi (kd) }{ \phi (d) } = \left ( 2^{b_2} \right ) \cdot \left ( 3^{b_3} \right ) \cdot \left ( 5^{b_5} \right ) \cdot \text{ ...}$

or
$\frac{ \phi (kd) }{ \phi (d) } = k$

Now, I thought that simplified expression to be kinda cool. But as I was thinking about it I checked:
$\frac { \phi (3 \cdot 4) }{ \phi (3) } = \frac{4}{2} = 2 \neq k$

So I've done something incorrectly. (Though I suspect the general structure of the proof is sound.)

Any thoughts?

-Dan
• Sep 27th 2013, 04:08 PM
emakarov
Re: Totient function and divisibility
The formula $\phi(p^k)=p^{k-1}(p-1)$ works only for k > 0.
• Sep 27th 2013, 04:40 PM
topsquark
Re: Totient function and divisibility
Quote:

Originally Posted by emakarov
The formula $\phi(p^k)=p^{k-1}(p-1)$ works only for k > 0.

Ah! So if a prime factor is missing...

I'll be back.

-Dan
• Sep 27th 2013, 07:42 PM
topsquark
Re: Totient function and divisibility
I think I've got the general idea, but I can't find a good way to express the final answer.

Let
$\large d = \prod _i p_i ^{a_i}$, where i runs over the primes in the decomposition.

$\large k = \prod _j p_j ^{b_j}$, ditto

$\large kd = \prod _{i, j} p_i^{a_i} p_j ^{b_j}$ i and j run over the decompositions and have no restrictions
(The above definitions solve the problem of having 0 exponenets.)

Continuing we have
$\large \phi (d) = \phi \left ( \prod _i p_i ^{a_i} \right )$

and
$\large \phi (kd) = \phi \left ( \prod _{i, j} p_i^{a_i} p_j ^{b_j} \right )$

We need to have some care about primes common to both factor sets. So break the factors into two groups: primes common to both k and d (x), and not shared (y):

$\large \phi (kd) = \phi \left $\left ( \prod _x p_x^{c_x} \right ) \left ( \prod _y p_y ^{e_y} \right ) \right$$

By construction the x product and y product are relatively prime, so we may factor:
$\large \phi (kd) = \phi \left ( \prod _x p_x^{c_x} \right ) \phi \left ( \prod _y p_y ^{e_y} \right )$

So this leaves us with
$\large \frac{\phi (kd) }{ \phi (d) } = \frac{ \phi \left ( \prod _x p_x ^{c_x} \right ) \phi \left ( \prod _y p_y ^{e_y} \right ) }{ \phi \left ( \prod _i p_i ^{a_i} \right ) }$

$\large = \frac{ \prod_x p_x ^{c_x - 1} (p_x - 1 )\cdot \prod_y p_y ^{e_y - 1} (p_y - 1) }{ \prod_i p_i ^{a_i - 1} ( p_i - 1) }$

Clearly the denominator is completely canceled by the factors in the numerator. But how do I show that? I'm a bit stumped on how to express this mathematically. (Perhaps because I need to get to bed!)

-Dan

PS Is this problem really this abstract (and long)? This is rather more intense than the other problems in the set.
• Sep 28th 2013, 07:28 AM
topsquark
Re: Totient function and divisibility
Success! All I needed to finish the proof came to me just after I woke up. :)

If that sounds nerdy I woke up at 3 AM the other night working on one point compactification in my sleep.

-Dan