let p be prime, show that p does not divide n where n is a positive integer if and only if phi(np) = (p-1)phi(n)
thanks in advance.
Whenever $\displaystyle (a,b)=1$ we have $\displaystyle \phi(ab)=\phi(a)\phi(b)$. So if $\displaystyle p \nmid n$, then $\displaystyle (p,n)=1$ and $\displaystyle \phi(pn)=\phi(p)\phi(n)=(p-1)\phi(n)$.
Conversely suppose $\displaystyle \phi(pn)=(p-1)\phi(n)$; write $\displaystyle n=p^\alpha m,\: p \nmid m$. Then $\displaystyle \phi(n)=\phi(p^\alpha)\phi(m)$ and $\displaystyle \phi(pn)=\phi(p^{\alpha+1})\phi(m)=\frac{\phi(p^{\ alpha+1})}{\phi(p^\alpha)}\phi(n)$. From this and our hypothesis we get $\displaystyle \frac{\phi(p^{\alpha+1})}{\phi(p^\alpha)} = p-1$. But if $\displaystyle \alpha>0$, $\displaystyle \frac{\phi(p^{\alpha+1})}{\phi(p^\alpha)} = \frac{p^{\alpha+1}(1-1/p)}{p^{\alpha}(1-1/p)} = p$, which contradicts $\displaystyle \frac{\phi(p^{\alpha+1})}{\phi(p^\alpha)} = p-1$; hence $\displaystyle \alpha=0$ and $\displaystyle p \nmid n$.