2

Prove or disprove:

Suppose that a and b are positive integer then φ($a^b$) = a$^{b-1}$ φ(a)

I try many positive integer and I think it is true statement but please any help how can I start to prove it if it is true statement ?!

I know that if a is prime then φ($a^b$) =$a^b$ - $a^{b-1}$ and φ(a)=a-1.

dr.rise
  • 619
  • 5
  • 15

2 Answers2

1

This follows directly from

Show that $\phi(mn) = \phi(m)\phi(n)\frac{d}{\phi(d)}$

For $m=n=a$ this gives $$\phi(a^2)=\phi(a)^2\cdot \frac{a}{\phi(a)}=\phi(a)a$$

Now use induction.

Dietrich Burde
  • 130,978
  • Thank you but I did not take any rule like $$\phi(mn) = \phi(m)\phi(n)\frac{d}{\phi(d)}$$ – dr.rise Dec 11 '17 at 10:21
  • Yes, perhaps you did not, but here on MSE it is a duplicate. The proof is easy, have a look. I do not think, any other proof for your question becomes much easier than this. – Dietrich Burde Dec 11 '17 at 10:23
1

Write $ a = \prod p_i^{e_i} $. Then

$ \phi(a^b) = \phi(\prod p_i^{be_i}) = \prod \phi(p_i^{be_i}) = \prod p_i^{be_i-1}(p_i-1) = \prod p_i^{be_i} \prod \frac{p_i-1}{p_i} = a^{b-1} a \prod \frac{p_i-1}{p_i} = a^{b-1} \phi(a) $

because $$ \phi(n)= n \prod_{p \mid n} \frac{p-1}{p} = n \prod_{p \mid n} \left(1-\frac{1}{p}\right) $$

Alternatively, consider $f(a) = \dfrac{\phi(a^b)}{a^{b-1}\phi(a)}$ and prove:

  • $f$ is multiplicative: If $\gcd(u,v)=1$, then $f(uv)=f(u)f(v)$.

  • $f(a)=1$ if $a$ is a prime power.

to conclude that $f(a)=1$ for all $a$.

lhf
  • 216,483