Wir berechnen ${\displaystyle {}\sigma (m)}$, ${\displaystyle {}\sigma (n)}$ und ${\displaystyle {}m+n}$. Es ist

${\displaystyle {}{\begin{aligned}\sigma (m)&=\sigma (2^{k}ab)\\&=\sigma (2^{k})\sigma (a)\sigma (b)\\&={\left(2^{k+1}-1\right)}{\left(3\cdot 2^{k-1}\right)}{\left(3\cdot 2^{k}\right)}\\&={\left(2^{k+1}-1\right)}\cdot 9\cdot 2^{2k-1}.\end{aligned}}}$

Weiter ist

${\displaystyle {}{\begin{aligned}\sigma (n)&=\sigma (2^{k}c)\\&=\sigma (2^{k})\sigma (c)\\&=(2^{k+1}-1)(1+c)\\&=(2^{k+1}-1)\cdot 9\cdot 2^{2k-1}.\end{aligned}}}$

Schließlich ist

${\displaystyle {}{\begin{aligned}m+n&=2^{k}(ab+c)\\&=2^{k}{\left({\left(3\cdot 2^{k-1}-1\right)}{\left(3\cdot 2^{k}-1\right)}+9\cdot 2^{2k-1}-1\right)}\\&=2^{k}{\left(9\cdot 2^{2k-1}-3\cdot 2^{k-1}-3\cdot 2^{k}+9\cdot 2^{2k-1}\right)}\\&=2^{k}{\left(9\cdot 2^{2k}-9\cdot 2^{k-1}\right)}\\&=2^{k}2^{k-1}\cdot 9{\left(2^{k+1}-1\right)}.\end{aligned}}}$