Problem
ALG-B3-M05-P010 Two Rules on Natural Numbers
#10
★★★☆☆ Level 3 of 5
A function \(f:\mathbb N\to\mathbb N\) satisfies \(f(mn)=f(m)f(n)\), \(f(n+1)=f(n)+2n+1\), and \(f(1)=1\). Find \(f(n)\).
The second condition already gives a recursion.
The recursion gives \(f(n+1)-f(n)=2n+1=(n+1)^2-n^2\). Since \(f(1)=1=1^2\), telescoping yields \(f(n)=n^2\). Multiplicativity checks: \(f(mn)=m^2n^2=f(m)f(n)\). The answer is \(f(n)=n^2\).
Mixes multiplicativity with a recursive Cauchy-type form.