This reduces to:
If $n \in \mathbb{N}$ s.t. $n$ has $n$ positive divisors, show $n = 1$ or $2$.
All I need to know is if the following is the correct way to think about the solution:
going backwards to get contradiction, we can look at $n \geq 3$
I appeal to "$n$ is prime or product of primes":
case(i) $n$ prime in which case the assumption will fail clearly
case (ii) $n$ product of primes, in this case the minimal number of divisors we get is 3 and that's with perfect squares.
Am I on the right track or not?
Best Answer
Here's a hint: notice that $\forall n \in \mathbb{N}, n$ and $n+1$ are relatively prime. Then if $d(n)=n$, every positive integer $1,...,n $ divides $n$. However, $n$ and $n-1$ must be coprimes (or relatively prime). From this, see if you can argue that the only $n$ satisfying this condition are $1$ and $2$.