Find all possible real numbers M and N that satisfy this system of simultaneous equations:
M + (3M-N)/(M2 + N2) = 3, and:
N = (M+3N)/(M2 + N2)
The second equation can be solved as M*N = 1/2 +/- sqrt[-N^4+3N^2+1/4].
Solving both equations for M^2+N^2 yields -(3M-N)/(M-3) = (M+3N)/N. This eventually solves to M = -3N+3/2 +/- sqrt[10N^2+9/4]
After lengthy algebra, three answers emerge: (M,N) = (2,1), (0,0), (1,-1). The (0,0) must be discarded for division by zero, leaving (2,1) and (1,-1) as the final answers.