0
$\begingroup$

Let $\phi (n)$ be the Euler totient function $n\in \Bbb N$. If $n$ has $r$ odd prime divisors then $2^r | \phi(n)$

If $n$ is prime then $\phi(n)= n-1$ except $2$ all primes are odd therefore $n-1$ would be even and $2^r$ a divisor.

What should I do if $n$ is not prime?

$\endgroup$
  • $\begingroup$ Perhaps the correct statement is "If $n$ has $r$ odd prime divisors, ....". $\endgroup$ – lhf May 11 '17 at 22:35
2
$\begingroup$

If $n$ is not prime, then $n=p_1^{\alpha_1}\cdots p_r^{\alpha_r}$, and $\phi(n)=n\prod_{i=1}^r\left(1-\frac{1}{p_i}\right)$. Rewriting the parenthetical as $\frac{p_i-1}{p_i}$,we see that $p_i$ will divide $n$ (since each $\alpha_i\geq 1)$ and there are $r$ terms of $p_i-1$ each of which is even (since $p_i$ are odd). Hence $2^r$ will divide the numerator.

$\endgroup$
  • $\begingroup$ We were given that there are $r$ odd prime divisors, so $2$ is not a problem. $\endgroup$ – Ross Millikan May 12 '17 at 5:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.