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September 18th, 2011, 02:31 AM  #1 
Joined: Sep 2011 Posts: 3 Thanks: 0  solve equation
determine all functions such that : , for each , hint please 
November 18th, 2011, 01:23 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 12,496 Thanks: 553 
Try giving x and y various values. For example, substituting x = f(0), y = f(0) into the equation gives f(f(0) + f(0)) = f(f(0) + f(0)) + (f(f(0)))². Hence f(f(0)) = 0. 
November 18th, 2011, 06:42 AM  #3 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 15,663 Thanks: 813 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Re: solve equation
I only get f(f(0)) = 0 from that. With x = y = 0 you can get two possibilities for f(0); eventually I was able to make one lead to contradiction leaving only f(0) = 0. Once you have that, take y = x to find f(2x) = f(x)  f(x)^2. If I had continuity I feel that this would give me a function for each choice of (say) f(1), but I don't have that. 
November 18th, 2011, 09:35 PM  #4 
Global Moderator Joined: Dec 2006 Posts: 12,496 Thanks: 553 
Thanks. I've corrected my previous post. Now for a better attempt . . . For any real u, substituting x = u + f(u), y = f(u) gives f(u + 2f(u)) = (f(u + f(u)))² + f(u + 2f(u)), so f(u + f(u)) = 0. Also, substituting x = u, y = 0 gives f(u + f(u)) = (f(u))² + f(u), so f(u)(f(u) + 1) = 0, i.e. f(u) = 0 or 1. If f(0) = 1, substituting x = 0, y = u in the equation gives f(f(u)) = 1 + f(u), which implies f(u) = 1 and f(f(u)) = 0, but that gives both f(1) = 1 and f(1) = 0 (by putting u = 1 and u = 0, respectively), a contradiction. Hence f(0) = 0. Putting y = x gives f(x + f(0)) = (f(x))² + f(2x), i.e. f(x) = (f(x))² + f(2x), which implies f(x) cannot be 1. Hence f(x) = 0 for all x. 

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