Prove that is even ifm> 2.

and

Find all integersnsuch that .

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, 09:57 AMkyldn6Combinatorial Study of Phi(n)
Prove that is even if

*m*> 2.

and

Find all integers*n*such that .

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, 10:36 AMtonio
- Nov 5th 2009, 07:33 AMharkapobi
- Nov 5th 2009, 08:06 AMchisigma
For the second question see here...

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

Kind regards

- Nov 5th 2009, 11:13 AMtonio
- Nov 5th 2009, 01:10 PMharkapobi
I'm sorry, i know i'm probably being really dumb but I still dont get it.

http://www.mathhelpforum.com/math-he...4ad3ebfd-1.gif

How do you find n? - Nov 5th 2009, 02:03 PMtonio

Check cases: is even unless , so you can have :

1)

The only form 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)

The only way we can have is if all the primes appear raised to 1, but then we must have , which is possible only if there's one single prime which then must equal 13.

So we already have , since

Now you try

Tonio - Nov 5th 2009, 02:54 PMharkapobi
Ok, I understand that I think. My question is actually

So I have

Case 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: where I get that there is a single prime 7. Then I also get n = 14 because

Case 3:

here I get 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) = 3 and this is not possible.

Case 4:

so there is one prime 3 raised to the power of 2 and all others raised to 1.

So (3-1) = 2 = 2 so there is only one other prime which is 2. Giving me = 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 :)