Euler's phi function: multiplicity

**glowplug19** · May 3rd 2009, 09:43 AM

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.

**glowplug19** · May 3rd 2009, 10:21 AM

Sorry, title should be "multiplicativity", not "multiplicity".

**TheAbstractionist** · May 3rd 2009, 11:13 AM

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 $\phi(2n)=\phi(n)+\phi(n)=2\phi(n).$

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