0
$\begingroup$

I wish to solve $\phi(n)=n$ for $n$ where $\phi(n)$ is the Euler's Totient function.

I solved this by splitting this into three possible cases:

($1$) $n$ is prime

($2$) $n$ is composite

($3$) $n=1$.

($1$) gives contradiction as $\phi(n)=n-1\not=n$.

($2$) gives contradiction as $\phi(n)\geq\frac{1}{2}n$.

($3$) Works and is the only left choice so we conclude that $n=1$.

Is there a nicer way to answer this question?

$\endgroup$
  • $\begingroup$ $\phi(n)\geq\frac{1}{2}n$ doesn't contradict $\phi(n) = n$. Besides, it isn't true (look at $n = 6$). We also don't have $\phi(n)\leq \frac12n$ (look at $n = 9$). $\endgroup$ – Arthur Mar 19 '18 at 7:22
3
$\begingroup$

There are only $n-1$ positive integers below $n$, so unless $\gcd(n, n) = 1$ we must have $\phi(n)\leq n-1$.

$\endgroup$
2
$\begingroup$

Note that $$n=\phi(n)=n\prod_p\left(1-\frac1p\right)=n\implies\prod_p\left(1-\frac1p\right)=1$$ which does not hold for any primes $p$, forcing $n=1$.

$\endgroup$

Your Answer

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