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.
Tomarken does a fine job of showing that n = p is a solution, and n = 2(p-1) is a solution for all primes p, but leaves out showing that these are the only solutions.
First, f(n) = 2 iff n is prime. If n is composite, then there’s at least one factor of n that’s >= 2 and < n, and so the smallest f(n) can be in this case is 3. So ONLY prime n’s can have f(n) = 2.
Now, if f(f(n)) = 2 then f(n) = some prime p > 2, and n = (p -1)k. We already know that when k = 1 we have a solution, so we can focus here on k > 1. We can similarly require p > 2 because if p = 2 then f(n) = 2 and we already know n must be prime.
If p is odd, then (p-1) can be written as 2q, and n = 2qk. But if k > 2 then kq is clearly a proper divisor of n AND kq > 2q, so f(n) would NOT actually be (p-1). This requires that k = 2 then. When k.= 2 then n = 2(p-1) and we have the only other solutions.
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Posted by Paul
on 2021-05-15 16:50:47 |
followup to solution
