TITLE: Relatively Prime Numbers and Number of Divisors of n
QUESTION [6 upvotes]: Find all positive integers $n$ such that $\phi(n) + \tau(n) > n$.
I'm not sure how to deal with such abstract functions... help?

REPLY [8 votes]: We are adding the numbers between $1$ and $n$ that satisfy the number divides $n$ or is relatively prime to $n$, notice at most $n$ numbers satisfy this and the number $1$ is the only one satisfying both these properties(so the max is $n+1$). So if there is a number between $1$ and $n$ that is neither relatively prime to $n$ or a divisor of $n$ we can conclude $\phi(n) + \tau(n) \leq n$.
If the number is composite and different to $4$ we can take the smallest prime $p$ dividing $n$ and then $n=pk$ with $k>2$. then we can change $k$ with $k-1$ and then $p(k-1)$ does not divide $pk$ and is not relatively prime to $n$.
Thus the only solutions are primes, $1$ and $4$.