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 3 of 6: Normalize the function
In plain words

Divide out the common factor.

g(t)=f(t)/k⇒g(n2g(m))=mg(n)2g(t)=f(t)/k\quad\Rightarrow\quad g(n^2g(m))=m g(n)^2
Detailed analysis

Put g=f/k g=f/k . The scaling identity shows g g is integer-valued, and substituting f(m)=kg(m) f(m)=kg(m) into the original equation gives g(n2g(m))=mg(n)2 g(n^2g(m))=m g(n)^2. If k>1 k>1, then g(1998)<f(1998) g(1998)<f(1998), so a minimizing function must have k=1 k=1.