Let R be the set of real numbers. Determine all functions f:R→R that satisfy the equation
f(x+f(x+y))+f(xy)=x+f(x+y)+yf(x)
for all real numbers x and y.
This is part of the set IMO 2015
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Comments
Let P(x,y) be the above FE.
P(0,0)⟹f(f(0))+f(0)=f(0)⟹f(f(0))=0 P(0,f(0))⟹2f(0)=f(f(0))+f(0)2⟹f(0)2−2f(0)=0⟹f(0)=0,2 Case 1: f(0)=0. Then P(0,y)⟹f(f(y))+f(0)=f(y)+yf(0)⟹f(f(y))=f(y) Since the range of f:R→R is R, then f(x)=x.
Case 2: f(0)=2. Then let g(x)=f(x)−2 which changes the FE to P(x):=g(g(x+y)+x+2)+f(xy)+2=x+g(x+y)+yg(x)+2y [I'll finish later]
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In case 1, you must prove that f is surjective to make that conclusion because R is the codomain of f, which is not necessarily the range.
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Oh right. I'll think about this problem some more.
It does look like f(x)=2−x satisfies the functional equation. My guess would be f(x)=x or f(x)=2−x, but I do agree with the comment that case 1 still has some work to be done to get to f(x)=x.
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Yes you are right @Patrick Corn !
f(x)=x or 2−x∀x
This problem is quite a bit harder than I perceived it to be. I don't think I can find a solution.
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Let P(x,y) be the assertion f(x+f(x+y))+f(xy)=x+f(x+y)+yf(x)
P(0,0)⇒f(f(0))=0P(0,f(0))⇒2f(0)=f(0)2⇒f(0)=0or2
case 1.
f(0)=2⇒f(2)=0P(x,1)⇒f(x+f(x+1))=x+f(x+1)P(0,f(x+1)+x)⇒f(x+1)+x+2=f(x+1)+x+2(f(x+1)+x)⇒f(x)=2−x∀∈R
Case 2.f(0)=0
P(x,0)⇒f(x+f(x))=x+f(x)P(x,1)⇒f(x+f(x+1))=x+1+f(x)P(1,f(x+1)+x)⇒f(1+f(1+x+f(x+1)))+f(x+f(x+1))=1+f(x+1+f(x+1))+f(x+1)+x⇒f(f(x)+x+1)=f(x)+x+1P(x,−1)⇒f(x+f(x−1))+f(−x)=x+f(x−1)−f(x)⇒−f(x)=f(−x)P(x,−x)⇒f(x)+f(−x2)=x−xf(x)P(−x,x)⇒f(−x)+f(−x2)=−x+x(−x)⇒f(x)−f(−x)=2x−x(f(x)+f(−x))⇒f(x)=x∀x∈R
hence, f(x)=2−x and f(x)=x are solutions.
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f(x+1)+x in the FE.
Oh dang, I never thought of putting something as unusual asLog in to reply
@Daniel Liu and that is why this is problem 5 of the IMO!
That is what most people never thought!Log in to reply
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However problems #1, #4 and #6 have been generally appreciated
It is not very hard to see that f(x) = x and f(x) = 2 - x are solutions. The difficult part is to show that they are the only solutions. In my eyes, Sualeh Asif demonstrates that f(0) = 2 implies that f(x) = x - 2. My compliments! But I am not convinced by his reasoning that f(0) = 0 implies that f(x) = x for all real x. In particular, I do not see the consequence of P(x,1) on line 2 of case 2. Neither the consequence of P(1,f(x+1)+x) on line 4 of case 2.