SOLUTION: If g(x) = x + x^2, prove that g(n) − 4g(m) = 0 has no solutions for positive integers m and n.

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Question 1162749: If g(x) = x + x^2, prove that g(n) − 4g(m) = 0 has no solutions for positive integers m and n.
Answer by ikleyn(52898) About Me  (Show Source):
You can put this solution on YOUR website!
.

            Obviously,  this problem is  EITHER  for advanced school (?) students  OR  for not-indifferent amateurs.

            Therefore,  I will give only the  idea  of the proof, without going into details. 


(1)  It is "almost obvious", that the best (most closest) approximation to 4g(m) by the numbers of the form  n+n^2 = n*(n+1)  is  g(2m).


     It can be checked by analyzing inequalities . . . 



(2)  At the same time, it is easy to check manually that  g(2m)  is not equal to 4g(m).


It is how I see the possible proof . . .