Problem
NT-B2-M10-P003 Multiplicativity of Sum of Divisors
#3
★★☆☆☆ Level 2 of 5
Prove that if \(\gcd(a,b)=1\), then \(\sigma(ab)=\sigma(a)\sigma(b)\).
Every divisor of \(ab\) is uniquely a product of a divisor of \(a\) and a divisor of \(b\).
Since \(a\) and \(b\) are coprime, every divisor \(d\mid ab\) is uniquely \(d=d_1d_2\), where \(d_1\mid a\), \(d_2\mid b\). Hence \(\sigma(ab)=\sum_{d_1\mid a}\sum_{d_2\mid b}d_1d_2=\left(\sum_{d_1\mid a}d_1\right)\left(\sum_{d_2\mid b}d_2\right)=\sigma(a)\sigma(b)\).
Standard proof by a divisor bijection.