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 September 18th, 2011, 02:31 AM #1 Newbie   Joined: Sep 2011 Posts: 3 Thanks: 0 solve equation determine all functions $f:R \rightarrow\ R$ such that : $f(x+f(x+y))=f(x-y)+f(x)^2$, for each $x,y \in R$, hint please
 November 18th, 2011, 01:23 AM #2 Global Moderator   Joined: Dec 2006 Posts: 15,341 Thanks: 1017 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: 16,046 Thanks: 931 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: 15,341 Thanks: 1017 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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