Find all functions f:R→R satisfying f(f(x+y))=f(x+y)+f(x)f(y)−xyfor all real numbers x and y.
Problem
Source: Indonesia IMO 2007 TST, Stage 2, Test 5, Problem 2
Tags: function, algebra proposed, algebra
15.11.2009 12:21
Raja Oktovin wrote: Find all functions f:R→R satisfying f(f(x+y))=f(x+y)+f(x)f(y)−xy for all real numbers x dan y. Let P(x,y) be the assertion f(f(x+y))=f(x+y)+f(x)f(y)−xy Let f(0)=a It's elementary first (just plug in the original equation) to see that the only solution of the type f(x)=bx+a is f(x)=x Subtracting P(x,y) from P(x+y,0), we get the new assertion Q(x,y) : af(x+y)=f(x)f(y)−xy Then Q(a,−a) ⟹ f(a)f(−a)=0 and so ∃u∈{f(−a),f(a)} such that f(u)=0 Then Q(x−u,u) ⟹ af(x)=−u(x−u) If a=0, Q(x,y) implies f(x)f(y)=xy and so either f(x)=x ∀x, which indeed is a solution, either f(x)=−x ∀x, which is not a solution. If a≠0, this implies f(x)=−uax+u2a=bx+a for some b and so (see beginning lines), f(x)=x, which is impossible since f(0)≠0 in this subcase. Hence the unique solution f(x)=x ∀x
15.11.2009 13:08
I have another solution with much more calculation the the one of Mr.pco. Let P(x,y) be the assertion f(f(x+y))=f(x+y)+f(x)f(y)−xy and let a:=f(0). We have P(x,0)⇒f(f(x))=(a+1)f(x),∀x∈R. Thus, let x=0 we have f(a)=a(a+1). Moreover, the assertion P(x,y) becomes: af(x+y)=f(x)f(y)−xy,∀x,y,∈R. Suppose that a≠0. We have: P(a,a)⇒af(2a)=a2(a+1)2−a2, yielding f(2a)=a2(a+2). P(2a,a)⇒af(3a)=a(a+1)a2(a+2)−2a2, yielding f(3a)=a2(a+1)(a+2)−2a=a(a3+3a2+2a−2). P(3a,a)⇒af(4a)=a(a+1)a(a3+3a2+2a−2)−3a2 P(2a,2a)⇒af(4a)=a4(a+2)2−4a2. This means that a(a+1)a(a3+3a2+2a−2)−3a2=a4(a+2)2−4a2⇔a2(a−3)(a+3)=0 But we have supposed that a≠0 then a=3 or a=−3. 1. If a=3 then P(x,y) becomes 3f(x+y)=f(x)f(y)−xy and hence f(3)=3⋅(3+1)=12. P(x,1)⇒3f(x+1)=f(1)f(x)−x. P(x+1,1)⇒3f(x+2)=f(1)f(1)f(x)−x3−(x+1) P(x+2,1)⇒3f(x+3)=f(1)f(1)f(1)f(x)−x3−(x+1)3−(x+2). P(x,3)⇒3f(x+3)=f(x)f(3)−3x=12f(x)−3x. Therefore, 12f(x)−3x=f(1)f(1)f(1)f(x)−x3−(x+1)3−(x+2),∀x∈R, contradiction. 2. If a=−3, the same argument yields another contradiction. Thus a=0 and we have f(x)f(y)=xy,∀x,y∈R. Put y=1 then f(x)=kx,∀x∈R, where k=1f(1). Now the assertion P(x,y) yields that k=1. Therefore, the only solution is f(x)=x.
27.12.2009 23:21
Hello!!!!!!
Roobiks [/hide]
28.12.2009 00:49
RobRoobiks wrote: Hello!!!!!! Let y=0 f(f(x))=f(x)+f(0)f(x) f(f(x))=(f(0)+1)f(x) thus its easy to see f(x)=(f(0)+1)x Yes, easy to see, but wrong. You got f(x)=(f(0)+1)x ∀x∈f(R) and not ∀x∈R (you did not show that f(x) is a surjection).
28.12.2009 16:35
Mashimaru wrote: Therefore, 12f(x)−3x=f(1)f(1)f(1)f(x)−x3−(x+1)3−(x+2),∀x∈R, contradiction. How do you reach the contradiction? For example if f(1)=0 then it is equivalent to f(x)=x−16. If f(1)≠3√108 then it is equivalent to f(x)=ax+b for some a,b∈R. If f(1)=3√108 then you can reach the contradiction at once. So you can only use it to see that f is a linear function. (Sorry if this was obvious)
28.12.2009 20:01
f(f(x+y))=f(x+y)+f(x)f(y)−xy (1) Denote f(0)=c. Take y=0 to get f(f(x))=f(x)(1+c). (2) Set y=−x in (1): f(f(0))−f(0)=x2−f(x)f(−x), but by (2), f(f(0))−f(0)=c2. Hence, f(−x)f(x)=c2−x2 (3). Take x=c in (3) to conclude there is a real number u satisfying f(u)=0. Take x=u in (2): f(0)=f(f(u))=f(u)(1+c)=0. It means f(f(x))=f(x), as c=0, but than from (1), f(x)f(y)=xy for all x,y. Easy to verify f(x)=0, for all x is not a solution. Take t such that f(t)=w and w is nonzero. wf(x)=xt, which can be rewritten as f(x)=kx, where k is a fixed real number. Plugging it in (1), (x+y)(k2−k)=xy(k2−1), for all x,y. If x=−y and they are both nonzero, the last equation would give k2−1=0. Assume k=−1; Than, for all pairs (x,y), 2(x+y)=0, which is clearly false. It means that k=1 and f(x)=x, which obviously satisfies the given conditions.
08.09.2023 15:32
Let P(x,y):=f(f(x+y))=f(x+y)+f(x)f(y)−xy P(0,0) yields f(f(0))=f(0)+f(0)2 P(−f(0),f(0)) yields f(f(0))=f(0)+f(−f(0))f(f(0))+f(0)2 from P(0,0)⟹f(0)+f(0)2=f(0)+f(0)2+f(−f(0))f(f(0))⟹f(f(0))f(−f(0))=0 Thus f(f(0))=0 or f(−f(0)) or both of them are equal to 0 Case 1: f(−f(0))=0 P(0,−f(0)) yields f(f(−f(0))=f(−f(0))+f(0)f(−f(0))⟹f(0)=0 Case 2: f(f(0))=0 Plugging this in P(0,0) we obtain f(0)+f(0)2=0⟹f(0)(f(0)+1)=0 thus either f(0)=0 or f(0)=−1 Subcase: f(0)=−1 This implies f(−1)=0 P(x,0) yields f(f(x))=f(x)+f(x)f(0)⟹f(f(x))=0 plugging this in P(x,y) yields f(x+y)+f(x)f(y)=xy:=Q(x,y) Q(x+1,−1) yields f(x)+f(x+1)f(−1)=−(x+1)⟹f(x)=−(x+1) however plugging this into P(x,y) yields a contradiction. Therefore since f(0)=0 in both of these cases we can conclude that f(0)=0 P(x,0) yields f(f(x))=f(x) plugging this in P(x,y) yields f(x+y)=f(x+y)+f(x)f(y)−xy⟹f(x)f(y)=xy:=T(x,y) T(x,1) yields f(x)f(1)=x⟹f(x)=xc where c=f(1) Pugging this in P(x,y) yields x+yc2=x+yc+xyc2−xy⟹(c−1)(x+y)=(c−1)xy⟹(c−1)(x+y−xy)=0 thus either c=1 or x+y=xy However the latter clearly yields a contradiction. Therefore c=1 which implies that f(x)=x In conclusion f(x)=x,∀x∈R ◼.