Hi can someone help me with this question,
"Give a brief argument to show that phi(n) is even for all n>2"
If $\displaystyle \gcd(a,b)=1$ then $\displaystyle \phi(ab)=1$. If $\displaystyle n>1$ and $\displaystyle p$ an odd prime divisor of it we can write $\displaystyle n=p^kj$ where $\displaystyle j$ not divisible by $\displaystyle p$. Then $\displaystyle \phi(n) = \phi(p^k)\phi(j)$. Thus, $\displaystyle \phi (p^k) = p^{k}-p^{k-1}$ which is even. Now finish the argument.
This is Mine 72th Post!!!