it isn't hard to solve the first equation for M and substitute it into the second. Set this equal to zero. A graph shows the solutions for (M,N) are (0,0) and (-1,-1). But it isn't clear these are the only ones.
If you instead expand the two LHSs and do the same as before it is easy to see the resulting equation as sort of like a polynomial of degree 6 except with some terms of fractional degree between 0 and 1. This graph shows the same two solutions and makes more clear there aren't any more.
Posted by Jer
on 2015-11-27 16:01:12