# euler fhi function and power of a prime

• Nov 29th 2011, 05:42 AM
seventhson
euler fhi function and power of a prime
if (n-1) is divisible by φ(n) then there is no prime p such that p^2|m.

Thanks...
• Nov 29th 2011, 11:54 AM
PaulRS
Re: euler fhi function and power of a prime
I suppose you meant to say $p^2 | n$ .

Let's proceed by contradiction, suppose on the contrary that there were a prime $p$ such that $p^2 | n$. Clearly $p\not | (n-1)$ since $n$ and $(n-1)$ are coprime, however $p^2 | n$ implies $p | \varphi ( n)$ $(*)$, and since $\varphi(n) | (n-1)$ we have that $p | (n-1)$ which is a contradiction. $\square$

$(*)$ Check this by using the formula $\phi(n) = n \cdot \prod_{p|n} \left( 1 - \frac{1}{p} \right)$ , where the products runs over all primes dividing $n$.
• Dec 1st 2011, 01:09 PM
seventhson
Re: euler fhi function and power of a prime
Thank you Paul for your nice proof...