Problem
ALG-B3-M06-P012 A Derivative on Natural Numbers
#12
★★★★☆ Level 4 of 5
Let \(f:\mathbb N\to\mathbb Z_{\ge0}\), \(f(mn)=mf(n)+nf(m)\), and \(f(p)=p\) for every prime \(p\). Find \(f(n)\).
Divide by \(n\): consider \(g(n)=\frac{f(n)}{n}\).
The condition gives \(g(mn)=g(m)+g(n)\). For a prime \(p\), \(g(p)=1\). If \(n=p_1^{a_1}\cdots p_k^{a_k}\), then \(g(n)=a_1+\cdots+a_k\). Hence \(f(n)=n(a_1+\cdots+a_k)\), that is, \(f(n)=n\Omega(n)\), where \(\Omega(n)\) is the number of prime factors counted with multiplicity.
Divisibility and factorisation in functional-equation form.