This is not a solution however as a substitute an investigation with exciting discoveries. We now have two cases: (1) n^2 - mn - m^2 = 1 or (2) n^2 - mn - m^2 = -1. Solving for n in both (1) and (2) yields n = (m + sqrt(5m^2 + four))/2 or n = (m + sqrt(5m^2 - four))/2. We can readily rule out the minus case (in the plus or minus for quadratics) considering those will not yield solutions for n among the many factors 1, 2, ... ,2007. Let's experiment with the 2 equations above for n and list probably the most options (m,n): (1, 1), (1,2), (2, three), (3, 5), (5,eight), (8, 13), (13, 21), ... This simply screams out Fibonacci sequence, so we conjecture that some (if now not all) of the solutions to (n^2 - mn - m^2)^2 = 1 are consecutive Fibonacci numbers. To prove this conjecture we let n = f_k and m = f_(k-1) the place f_n is the nth Fibonacci number with f_0 = 0 and f_1 = 1. Take into account that f_k = f_(k-1) + f_(k-2) for ok >= 2. We want to show that (f_k)^2 - (f_k)(f_(k-1)) - (f_(ok-1))^2 ^2 = 1 it's a good recreation to prove that yourself if you haven't noticeable it earlier than. It turns out that the expression (f_k)^2 - (f_k)(f_(okay-1)) - (f_(ok-1))^2 is the identity f_(ok+1)*f_(ok-1) - (f_k)^2 = (-1)^okay in cover (whenever you tweak it a little bit). After we rectangular it the correct hand side becomes 1 as preferred. The evidence strongly suggests that the Fibonacci numbers supply us the only method to the eqn. (n^2 - mn - m^2)^2 = 1 over the nonnegative integers. Nevertheless, i am not certain the best way to exhibit this. If any individual else can exhibit that (or disprove it ;)) that will be exceptional. Assuming consecutive Fibonacci numbers give us the one options, then we comfortably must to find the most important Fibonacci quantity that is lower than or equal to 2007 which is 1597: that is n. Then, m is the preceding Fibonacci number, 987. These must provide us the max worth of m^2 + n^2. Once more, it will be a first-rate aid if any one could fill in the blanks in my work. *Addition (after seeing Duke's proof)* thanks so much for posting the proof! Lamentably, i know very little about Diophantine Eqns so one of the crucial stuff went over my head, but I recognize it nevertheless.