euler phi-function

**htata123** · March 29th 2009, 11:00 AM

Hi,

I need help with this proof:

Prove that there are no integers n for which phi(n) = n/4

Appreciate any assistance

**ThePerfectHacker** · March 29th 2009, 11:10 AM

Originally Posted by htata123

Hi,

I need help with this proof:

Prove that there are no integers n for which phi(n) = n/4

Appreciate any assistance

Write $n$ in a prime factorization and use formula for the phi-function.

**htata123** · March 29th 2009, 11:13 AM

I did that but I'm still stuck. Any ideas on how I should proceed?

**ThePerfectHacker** · March 29th 2009, 07:42 PM

Originally Posted by htata123

I did that but I'm still stuck. Any ideas on how I should proceed?

Certainly $n$ is divisible by $4$ so we can write $n=2^a\cdot m$ where $a\geq 2$ and $m$ is a positive odd integer. Now, $\phi(n) = \phi(2^a)\phi(m) = 2^{a-1}\phi(m)$ since $(2^a,m)=1$ . While $n/4 = 2^{a-2}m$ . Therefore, if $\phi(n) = n/4$ then $2^{a-1}\phi(m) = 2^{a-2}m\implies 2\phi(m) = m$ . This is impossible because LHS is even while RHS is odd.

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