Find the number of ordered pairs \((m, n)\) of integers that satisfy
\(mn = 2m + 4n\)
\(\begin{array}{|rcll|} \hline \mathbf{mn} &=& \mathbf{2m + 4n} \\ mn-2m - 4n &=& 0 \\ m(n-2) - 4n &=& 0 \\ m(n-2) - 4n +8 &=& 8 \\ m(n-2) - 4(n-2) &=& 8 \\ \mathbf{(n-2)(m-4)} &=& \mathbf{8} \\ \hline \end{array}\)
4 solutions:
\(\begin{array}{|rcll|} \hline 1) & 8 &=& 1*8 \\ 2) & 8 &=& 2*4 \\ 3) & 8 &=& 4*2 \\ 4) & 8 &=& 8*1 \\ \hline \end{array}\)
\(\begin{array}{|rcll|} \hline 1) & (n-2)(m-4) &=& 1*8 \\ & n-2 =1 && m-4 = 8 \\ & \mathbf{n =3} && \mathbf{m = 12} \\ \hline \end{array} \begin{array}{|rcll|} \hline 2) & (n-2)(m-4) &=& 2*4 \\ & n-2 =2 && m-4 = 4 \\ & \mathbf{n =4} && \mathbf{m = 8} \\ \hline \end{array} \\ \begin{array}{|rcll|} \hline 3) & (n-2)(m-4) &=& 4*2 \\ & n-2 =4 && m-4 = 2 \\ & \mathbf{n =6} && \mathbf{m = 6} \\ \hline \end{array} \begin{array}{|rcll|} \hline 4) & (n-2)(m-4) &=& 8*1 \\ & n-2 =8 && m-4 = 1 \\ & \mathbf{n =10} && \mathbf{m = 5} \\ \hline \end{array}\)
