A positive proper divisor is a positive divisor of a number, excluding itself. For positive integers n≥2, let f(n) denote the number that is one more than the largest proper divisor of n. Determine all positive integers n such that f(f(n)) = 2.
Yes, well Paul also does a fine job proving tomarken's solutions are the complete set. I did this too, but I think here explanation is a little simpler (i.e., no k). In fact, I didn't post at first because I took it to be what tomarken meant:
Clearly all n=p work, since, as t said, f(f(p))=f(2)=2.
Next consider all candidate n>3 which are not prime:
As P says: f(n)=2 iff n=p. Since f(f(n))=2, f(n)=p.
But f(n)-1 is both even and the greatest divisor of a number, and so must be the greatest divisor of an even number, so it is n/2. Substituting p for f(n) here gives p-1=n/2 and thus the remaining set of solutions: n=2(p-1)
Edited on May 16, 2021, 6:14 pm
followup to followup
