1. Okay I was thinking it over and you are right I don't think I made any errors. I don't know why I used n/2. I think I was just in a hurry to write it out lol

2. Originally Posted by BBAmp
Okay I was thinking it over and you are right I don't think I made any errors. I don't know why I used n/2. I think I was just in a hurry to write it out lol

Can anyone else confirm his results?

3. Suppose $n$ is even and $n=2^am$ where $m$ is odd.

$\phi(2n)=\phi(2^{a+1}m)=\phi(2^{a+1})\phi(m) = 2^a\phi(m)$

$2\phi(n)=2\phi(2^am)=2\phi(2^a)\phi(m)=2\cdot2^{a-1}\phi(m)=2^a\phi(m)$

Thus $\phi(2n)=2^a\phi(m)$ and $2\phi(n)=2^a\phi(m)\implies \phi(2n)=2\phi(n)$

4. Originally Posted by chiph588@
Suppose $n$ is even and $n=2^am$ where $m$ is odd.

$\phi(2n)=\phi(2^{a+1}m)=\phi(2^{a+1})\phi(m) = 2^a\phi(m)$

$2\phi(n)=2\phi(2^am)=2\phi(2^a)\phi(m)=2\cdot2^{a-1}\phi(m)=2^a\phi(m)$

Thus $\phi(2n)=2^a\phi(m)$ and $2\phi(n)=2^a\phi(m)\implies \phi(2n)=2\phi(n)$

Okay so is his case for when n is odd valid?

5. Originally Posted by 1337h4x
Okay so is his case for when n is odd valid?
This is for when n is even

6. Originally Posted by chiph588@
This is for when n is even
I know, but earlier a case was explained for when n was odd. Was the explanation that ensued sufficient and correct?

7. Originally Posted by 1337h4x
I know, but earlier a case was explained for when n was odd. Was the explanation that ensued sufficient and correct?
The explanation I gave for when n is odd is a valid proof.

8. Thank you!

Page 2 of 2 First 12

,

,

,

,

,

,

,

,

,

,

phi of 2n is equal to phi of n

Click on a term to search for related topics.