well if the answer is more than 3, I would be surprised. There are probably some integer rules to justify really small limits, but here is how I justified it to myself:
I. Treat every variable like it is a. 2a^2 <= 4a , or a^2 <= 2a . This only works for a=1, a=2. So likely a cannot be more than 2. a) suppose a=1 and treat d as a c. b+c^2 = 1+b+2c c^2 = 1+2c c^2 - 2c - 1 = 0 c^2 - 2c +1 = 2 (c-1)^2 = 2 c = 1+- sqrt(2) hmm.. so c doesnt even reach 3 b) suppose a=2 and treat d as a c. 2b + c^2 = 2 + b + 2c b = -c^2 - 2c + 2 with values of c>=3 , the right side is negative. II. conclusions? a is at most 2, and c is at most 2. So yea, basically what Don said. 1, 1, 2, 3 1, 2, 2, 3 2, 2, 2, 2 On Aug 30, 5:05 pm, Don <dondod...@gmail.com> wrote: > There are three solutions, with d never exceeding 3. > Don > > On Aug 30, 3:08 pm, him <himanshuarora.1...@gmail.com> wrote: > > > > > > > > > finding number of integral solution of the equation > > > ab+cd=a+d+b+c (1<=a<=b<=c<=d) (any efficient method ) -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
