Euler's phi function: multiplicity

• May 3rd 2009, 10:43 AM
glowplug19
Euler's phi function: multiplicity
I am supposed to show that, for Euler's totient function:

a.) If n is an odd integer, then $\phi(2n)=\phi(n)$.
and
b.) If n is an even integer, then $\phi(2n)=2\phi(n)$.

I thought I would use the multiplicity of phi to establish this, i.e.
$\phi(2n)=\phi(2)\cdot\phi(n)=1\cdot\phi(n)=\phi(n)$.

Obviously this doesn't work, because while it works for (a), it contradicts (b). Not sure why.
• May 3rd 2009, 11:21 AM
glowplug19
Sorry, title should be "multiplicativity", not "multiplicity".
• May 3rd 2009, 12:13 PM
TheAbstractionist
Quote:

Originally Posted by glowplug19
I am supposed to show that, for Euler's totient function:

a.) If n is an odd integer, then $\phi(2n)=\phi(n)$.
and
b.) If n is an even integer, then $\phi(2n)=2\phi(n)$.

I thought I would use the multiplicity of phi to establish this, i.e.
$\phi(2n)=\phi(2)\cdot\phi(n)=1\cdot\phi(n)=\phi(n)$.

Obviously this doesn't work, because while it works for (a), it contradicts (b). Not sure why.

Hi glowplug19.

Be careful here. The formula $\phi(mn)=\phi(m)\phi(n)$ only works when $\gcd(m,n)=1.$ This is why it works for (a) but not for (b), as $\gcd(2,n)=1$ if and only if $n$ is odd.

For (b) consider $S_1=\{1,2\ldots,n\}$ and $S_2=\{n+1,n+2\ldots,n+n=2n\}.$ If $n$ is even, the integers in $S_1$ which are coprime with $n$ excludes the even integers; hence an integer in $S_1$ is coprime with $2n$ if and only if it is coprime with $n.$ For $S_2$ note that $n+r$ is coprime with $n$ if and only if $r$ is coprime with $n;$ hence the same argument can be applied to $S_2$ to show that there are $\phi(n)$ integers in $S_2$ coprime with $2n.$ Thus $\phi(2n)=\phi(n)+\phi(n)=2\phi(n).$