MathLabs

Problem 6

Determine the least possible value of f(1998) f(1998), where f:N→N f:\mathbb{N}\to\mathbb{N} is a function such that for all m,n∈N m,n\in\mathbb{N}, f(n2f(m))=m(f(n))2. f\left(n^{2}f(m)\right)=m\left(f(n)\right)^{2}.
Step 4 of 6: Classify normalized solutions
In plain words

The normalized equation forces an involutive multiplicative map.

f(f(t))=t,f(st)=f(s)f(t)f(f(t))=t,\quad f(st)=f(s)f(t)
Detailed analysis

With f(1)=1 f(1)=1, the identities become f(f(t))=t f(f(t))=t and f(t2)=f(t)2 f(t^2)=f(t)^2. Applying the original equation with m=f(t2) m=f(t^2) gives f(st)2=f(s)2f(t)2 f(st)^2=f(s)^2f(t)^2, hence positivity gives multiplicativity.