Please hint me.
Let $S$ be the number non trivial divisors $n$. prove that $ S<\phi(n)+1$. $\phi$ is euler function.
hint: we know $ \sum_{d|n}\phi(d)=n$ so $ \phi(1)+\phi(n)+\sum_{1,n\not=d|n}\phi(d)=n$, thus $\phi(n)+1=n-\sum_{1,n\not=d|n}\phi(d)$.
thanks in advance.
Let $D(n)$ denote the number of divisors of $n \in \Bbb N^+$. Then it is not hard to show that $D$, like $\phi(n)$ is a multiplicative function (if $\gcd(m,n)=1$ then $D(mn)=D(m)D(n))$. Note that $D(p^a)=a+1$ for each prime $p$. So if we compare $D(p^a)$ with $\phi(p^a)=ap^{a-1}$ we see that for almost all $n \in \Bbb N^+$ we have $D(n)<\phi(n)$. The values for which $\phi(n)-S(n)\leq 0$ are $1, 2, 3, 4, 6, 8, 10, 12, 18, 24, 30$ with values for the difference $ 0, -1, 0, -1, -2, 0, 0, -2, 0, 0, 0$ respectively so that $\forall \in \Bbb N^+$ we have $\phi(n)+3-S(n) > 0$.
