Euler totient related identities/questions

Question

(a) Show that if $q$ is a prime and $q\ |\ m$ then $\phi(mq)=q\phi(m)$, while if $q \nmid m$ then $\phi(mq) = (q-1) \phi(m)$. Deduce if $m|n$ then $\phi(m)\ |\ \phi(n)$.

(b) If $\phi(n)=8$ and $p$ is a prime divisor of $n$, show that $p\leq5$. Hence find all $n$ such that $\phi(n)=8$.

Not sure where to start. I can't see how to use the formula $\phi(m)=m\prod_{i=1}^{k} \left(1-\frac{1}{p_i}\right)$ when $m= \prod_{i=1}^{k} p_i^{e_i}$.

(c) Show that for no integer $n$ is $\phi(n) = 14$.

Again I need a hint.

first prove that $\phi(q^k) = q^k-q^{k-1}$, then $\phi(m q^k) = \phi(m) \phi(q^k)$ when $gcd(m,q^k) = 1$ — reuns, Feb 20 '16 at 01:45

score 1 · Answer 1 · answered Feb 19 '16 at 22:18

1

The deduce bit in part (a) is not right. For in general $\varphi(cm)\ne c\varphi(m)$. For a proof, I would suggest strong induction, using the result of the first part of (a).

For (b), note that if $p$ is a prime that divides $n$, then $p-1$ divides $\varphi(n)$. If $p=7$, then $p-1$ does not divide $8$, and if $p\gt 7$, then $p-1\gt 8$, so again does not divide $8$.

The fact that $14$ is not $\varphi(n)$ for any $n$ has been proved a number of times on MSE.

answered Feb 19 '16 at 22:18

André Nicolas

514,336

Thanks, for (a) is there another way? I feel as if strong induction would be quite long for a deduce question For part (b) - the second part. I have $p=2,3,5$ so we can write $n=2^a \times 3^b \times 5^c$ Using the Eulier totient formula we have 7 ways to pick primes. But 2 of them will – kingofmathslads Feb 19 '16 at 23:21
The divisibility follows immediately from the fromula that says that $\varphi(\prod p_i^{a_i)}=\prod (p_i-1)p_i^{a_i-1}$, but then we don't use the first part. As to (b), clearly $b\le 1$ and $c\le 1$. Then we have cases $b=1$, $c=1$ (2 possibilities, for we can multiply by $2$). Also $b=1$, $c=0$. Then we need $a=3$. Aso $b=0$, $c=1$. Then $a=2$. Finally, $b=c=0$, $a=4$. – André Nicolas Feb 19 '16 at 23:34
why is b≤1 and c≤1 – kingofmathslads Feb 19 '16 at 23:57
Because if for example $b\ge 2$, then $\varphi(n)$ will be divisible by $3$. But $8$ is not divisible by $3$. – André Nicolas Feb 20 '16 at 00:00
why is $\phi(n)$ divided by 3 – kingofmathslads Feb 20 '16 at 00:29
Because $\varphi(3^k)=(2)(3^{k-1})$. – André Nicolas Feb 20 '16 at 00:36
Could I ask how you got the values for a? – George Feb 25 '16 at 01:24
1

It depends on what $b$ and $c$ are. We use the fact that if $a\ge 1$ then $\varphi(2^a)=2^{a-1}$. – André Nicolas Feb 25 '16 at 01:26
If b=1 and c=0. Why do we need a=3? – George Feb 25 '16 at 01:37
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Because $\varphi(3)=2$. So to get to $\varphi(n)=8$, we need $\varphi(2^a)=4$, and therefore we need $2^a=8$. – André Nicolas Feb 25 '16 at 01:42

Euler totient related identities/questions

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