Mamy równanie

$x^3\left(m-2\right)x^2+\left(2n+1\right)x-1=0$  

Rozważmy wielomian lewej strony

$W\left(x\right)=x^3\left(m-2\right)x^2+\left(2n+1\right)x-1$  

Jeśli W ma mieć pierwiastki całkowite to muszą one być dzielnikami wyrazu wolnego. Stąd są to liczby 1 i -1. Zatem

$W\left(1\right)=0\text{i}W\left(-1\right)=0$  

  $\{\begin{array}{l}1+m-2+2n+1-1=0\\ -1+m-2-2n-1-1=0\end{array}$  

  $\{\begin{array}{l}m-1+2n=0\mid +1-2n\\ m-2n-5=0\end{array}$  

  $\{\begin{array}{l}m=1-2n\\ m-2n-5=0\end{array}$  

  $\{\begin{array}{l}m=1-2n\\ 1-2n-2n-5=0\end{array}$  

  $\{\begin{array}{l}m=1-2n\\ -4n-4=0\mid +4\end{array}$  

  $\{\begin{array}{l}m=1-2n\\ -4n=4\mid :\left(-4\right)\end{array}$  

  $\{\begin{array}{l}m=1-2n\\ n=-1\end{array}$  

  $\{\begin{array}{l}m=1-2\cdot \left(-1\right)\\ n=-1\end{array}$  

  $\{\begin{array}{l}m=1+2\\ n=-1\end{array}$  

  $\{\begin{array}{l}m=3\\ n=-1\end{array}$