Find the number of ordered pairs (m, n) of integers that satisfy
mn = 2m + 4n
+202 Seeing as we have a Diophantine Equation with mn, m, and n, it's probably a good idea to use Simon's Favorite Factoring Trick. Move all the terms to one side:
\(mn-2m-4n=0\)
Factor out an m from the first two terms:
\(m(n-2)-4n=0\)
I will leave it to you to figure out what to do next.
Hint: Try to find a way to get another (n-2). You already have an -4n.
a=1; b=1;c=a*b; if(c==2*a+4*b, goto4, goto5);printc, a, b; a++;if(a<100, goto2, 0);a=1;b++;if(b<100, goto2, discard=0;
mn m n
36 12 3
32 8 4
36 6 6
50 5 10
+26393 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:
\(\begin{array}{|rcll|} \hline 1) & 8 &=& 1*8 \\ 2) & 8 &=& 2*4 \\ 3) & 8 &=& 4*2 \\ 4) & 8 &=& 8*1 \\ \hline \end{array}\)
\(\begin{array}{|rcll|} \hline 1) & (n-2)(m-4) &=& 1*8 \\ & n-2 =1 && m-4 = 8 \\ & \mathbf{n =3} && \mathbf{m = 12} \\ \hline \end{array} \begin{array}{|rcll|} \hline 2) & (n-2)(m-4) &=& 2*4 \\ & n-2 =2 && m-4 = 4 \\ & \mathbf{n =4} && \mathbf{m = 8} \\ \hline \end{array} \\ \begin{array}{|rcll|} \hline 3) & (n-2)(m-4) &=& 4*2 \\ & n-2 =4 && m-4 = 2 \\ & \mathbf{n =6} && \mathbf{m = 6} \\ \hline \end{array} \begin{array}{|rcll|} \hline 4) & (n-2)(m-4) &=& 8*1 \\ & n-2 =8 && m-4 = 1 \\ & \mathbf{n =10} && \mathbf{m = 5} \\ \hline \end{array}\)
