Prove that $\displaystyle \phi(m)$ is even ifm> 2.

and

Find all integersnsuch that $\displaystyle \phi(n) = 12$.

For the second one is it possible to do it simply with inspection or is there a formula I'm just missing?

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- Oct 29th 2009, 08:57 AMkyldn6Combinatorial Study of Phi(n)
Prove that $\displaystyle \phi(m)$ is even if

*m*> 2.

and

Find all integers*n*such that $\displaystyle \phi(n) = 12$.

For the second one is it possible to do it simply with inspection or is there a formula I'm just missing? - Oct 29th 2009, 09:36 AMtonio

If $\displaystyle m=\prod\limits_{n=1}^r p_i^{a_i}$ $\displaystyle \,,\,\,p_i\,\,primes\,,\,\,0<a_i\in \mathbb{N}$, then

$\displaystyle \phi(m)=m\,\prod\limits_{n=1}^r\left(1-\frac{1}{p_i}\right)$

This answers question 1 at once, and for question 2 you have a little maths to do. Enjoy!(Wink)

Tonio - Nov 5th 2009, 06:33 AMharkapobi
- Nov 5th 2009, 07:06 AMchisigma
For the second question see here...

http://www.mathhelpforum.com/math-he...t-problem.html

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$ - Nov 5th 2009, 10:13 AMtonio

I'll give you a little additional hint: if $\displaystyle n=p_1^{r_1}\cdot...\cdot p_k^{r_k}\,,\,\,then\,\,\phi(n)=p_1^{r_1}\cdot...\ cdot p_k^{r_k}\left(1-\frac{1}{p_1}\right)\cdot...\cdot \left(1-\frac{1}{p_k}\right)$ $\displaystyle =p_1^{r_1-1}\cdot...\cdot p_k^{r_k-1}(p_1-1)\cdot...\cdot (p_k-1)$.

OTOH, we know $\displaystyle 12=2^2\cdot 3$, so...

Tonio - Nov 5th 2009, 12:10 PMharkapobi
I'm sorry, i know i'm probably being really dumb but I still dont get it.

$\displaystyle 2^{2}3$ http://www.mathhelpforum.com/math-he...4ad3ebfd-1.gif

How do you find n? - Nov 5th 2009, 01:03 PMtonio

Check cases: $\displaystyle p_i-1$ is even unless $\displaystyle p_i=2$, so you can have $\displaystyle 12=12\cdot 1=6\cdot 2=4\cdot 3$:

1) $\displaystyle 12=12\cdot 1=\left[p_1^{r_1-1}\cdot...\cdot p_k^{r_k-1}\right]\left[(p_1-1)\cdot...\cdot (p_k-1)\right]$

The only form $\displaystyle (p_1-1)\cdot...\cdot (p_k-1)$ can be 1 is if there's only one prime and it is 2, but then the prime 3 lacks us, so this can't be:

2) $\displaystyle 12=1\cdot 12=\left[p_1^{r_1-1}\cdot...\cdot p_k^{r_k-1}\right]\left[(p_1-1)\cdot...\cdot (p_k-1)\right]$

The only way we can have $\displaystyle 1 =\left[p_1^{r_1-1}\cdot...\cdot p_k^{r_k-1}\right]$ is if all the primes appear raised to 1, but then we must have $\displaystyle 12 = \left[(p_1-1)\cdot...\cdot (p_k-1)\right]$ , which is possible only if there's one single prime which then must equal 13.

So we already have $\displaystyle \phi(13)=12\Longrightarrow \phi(2*13)=\phi(26)=12$ , since $\displaystyle \phi(2)=1$

Now you try $\displaystyle 6\cdot 2\,,\,\,2\cdot 6\,,\,\,4\cdot 3\,\,and\,\,3\cdot 4$

Tonio - Nov 5th 2009, 01:54 PMharkapobi
Ok, I understand that I think. My question is actually $\displaystyle \phi(n) = 6$

So I have

Case 1: $\displaystyle 6\cdot 1$ where http://www.mathhelpforum.com/math-he...71b264d4-1.gif can only be 1 if there is only 1 prime and its 2. But then no 3 to make 6.

Case 2: $\displaystyle 1\cdot 6$ where I get that there is a single prime 7. Then I also get n = 14 because $\displaystyle \phi(2\cdot 7) = 1\cdot 6$

Case 3: $\displaystyle 2\cdot 3$

here I get $\displaystyle \left[p_1^{r_1-1}\cdot ... \cdot p_k^{r_k-1}\right] = 2$ so then there is one prime 2 raised to the power of 2, then all others raised to the power of 1.

But http://www.mathhelpforum.com/math-he...71b264d4-1.gif = (2 - 1) $\displaystyle \cdot ... (p_k - 1)$ = 3 and this is not possible.

Case 4: $\displaystyle 2\cdot 3$

$\displaystyle \left[p_1^{r_1-1}\cdot ... \cdot p_k^{r_k-1}\right] = 3$ so there is one prime 3 raised to the power of 2 and all others raised to 1.

So (3-1)$\displaystyle \cdot ... (p_k - 1)$ = 2 $\displaystyle \cdot ... (p_k - 1)$ = 2 so there is only one other prime which is 2. Giving me $\displaystyle 3^2\cdot 2$ = 18

but i know i'm supposed to get a 9 from somewhere. I can't see where tho. Sorry to be a pain. Thank you for all the help.

Katy :)