## Lehmer Mack 2

Expand Messages
• Consider phi(n^2)=k.(n-1) Obviously k contains some factor of n-1 in it s denominator. Therefore, as phi(n^2)=n.phi(n): phi(n)=(k/n).(n-1) And therefore k/n is
Message 1 of 3 , Apr 2, 2002
Consider phi(n^2)=k.(n-1)

Obviously k contains some factor of n-1 in it's denominator.

Therefore, as phi(n^2)=n.phi(n):

phi(n)=(k/n).(n-1)

And therefore k/n is never an integer.

Jon Perry
perry@...
http://www.users.globalnet.co.uk/~perry/maths
BrainBench MVP for HTML and JavaScript
http://www.brainbench.com
• ... Let n=2 Phi(2^2) = 2 = 2.(2-1) = k=2 = k/n = 2 Now, as someone with limited intelligence , I think that 2 actually _is_ an integer, but am prepared to
Message 2 of 3 , Apr 2, 2002
--- Jon Perry <perry@...> wrote:
> Consider phi(n^2)=k.(n-1)
>
> Obviously k contains some factor of n-1 in it's denominator.
>
> Therefore, as phi(n^2)=n.phi(n):
>
> phi(n)=(k/n).(n-1)
>
> And therefore k/n is never an integer.

Let n=2
Phi(2^2) = 2 = 2.(2-1)
=> k=2
=> k/n = 2

Now, as someone with "limited intelligence", I think that 2 actually
_is_ an integer, but am prepared to be proved wrong using your Most
Valued mathematical techniques. After you've read Matthew 7 verse 5,
that is.

Phil

__________________________________________________
Do You Yahoo!?
Yahoo! Tax Center - online filing with TurboTax
http://taxes.yahoo.com/
• Well spotted. I had confused the numerator/denominator. ... and also a prime! ... Don t read the Bible. This verse states? Jon Perry perry@globalnet.co.uk
Message 3 of 3 , Apr 3, 2002
Well spotted. I had confused the numerator/denominator.

> I think that 2 actually
>_is_ an integer

and also a prime!

>After you've read Matthew 7 verse 5,that is.

Don't read the Bible. This verse states?

Jon Perry
perry@...
http://www.users.globalnet.co.uk/~perry/maths
BrainBench MVP for HTML and JavaScript
http://www.brainbench.com

-----Original Message-----
From: Phil Carmody [mailto:thefatphil@...]
Sent: 03 April 2002 00:03
Subject: Re: [PrimeNumbers] Lehmer Mack 2

--- Jon Perry <perry@...> wrote:
> Consider phi(n^2)=k.(n-1)
>
> Obviously k contains some factor of n-1 in it's denominator.
>
> Therefore, as phi(n^2)=n.phi(n):
>
> phi(n)=(k/n).(n-1)
>
> And therefore k/n is never an integer.

Let n=2
Phi(2^2) = 2 = 2.(2-1)
=> k=2
=> k/n = 2

Now, as someone with "limited intelligence", I think that 2 actually
_is_ an integer, but am prepared to be proved wrong using your Most
Valued mathematical techniques. After you've read Matthew 7 verse 5,
that is.

Phil

__________________________________________________
Do You Yahoo!?
Yahoo! Tax Center - online filing with TurboTax
http://taxes.yahoo.com/

Unsubscribe by an email to: primenumbers-unsubscribe@egroups.com
The Prime Pages : http://www.primepages.org

Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
Your message has been successfully submitted and would be delivered to recipients shortly.