MathLabs

Problem 3

Let N\mathbb{N} denote the set of positive integers. A function f:N→Nf:\mathbb{N}\to\mathbb{N} is said to be bonza if f(a)f(a) divides ba−f(b)f(a)b^a-f(b)^{f(a)} for all positive integers aa and bb. Determine the smallest real constant cc such that f(n)≤cnf(n)\le cn for all bonza functions ff and all positive integers nn.
Step 4 of 7: f(n) is always a power of 2
f(n) is a power of 2 for every n,f(n)=1 for odd nf(n)\ \text{is a power of }2\ \text{for every }n,\quad f(n)=1\ \text{for odd }n
Detailed analysis

Suppose an odd prime pp divides f(n)f(n) for some nn; then (assuming f≠idf\ne\mathrm{id}, the only case left to bound) P(n,p)P(n,p) gives f(n)∣pn−f(p)f(n)=pn−1f(n)\mid p^n-f(p)^{f(n)}=p^n-1 using step 3, so in particular p∣pn−1p\mid p^n-1, which is false since p∣pnp\mid p^n. Hence no odd prime divides f(n)f(n), i.e. f(n)f(n) is a power of 22 for every nn. Combined with step 1 (f(n)∣nnf(n)\mid n^n), if nn is odd then nnn^n is odd, so the power of 22 dividing it must be 11, giving f(n)=1f(n)=1 for all odd nn.