Problem 5! IMO 2015

Let R\mathbb{R} be the set of real numbers. Determine all functions f:RRf:\mathbb{R}\to\mathbb{R} that satisfy the equation

f(x+f(x+y))+f(xy)=x+f(x+y)+yf(x)f(x+f(x+y))+f(xy)=x+f(x+y)+yf(x)

for all real numbers xx and yy.

This is part of the set IMO 2015

Note by Sualeh Asif
4 years, 3 months ago

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Let P(x,y)P(x,y) be the above FE.

P(0,0)    f(f(0))+f(0)=f(0)    f(f(0))=0P(0,0)\implies f(f(0))+f(0)=f(0)\implies f(f(0))=0 P(0,f(0))    2f(0)=f(f(0))+f(0)2    f(0)22f(0)=0    f(0)=0,2P(0, f(0))\implies 2f(0)=f(f(0))+f(0)^2\implies f(0)^2-2f(0)=0\implies f(0)=0, 2 Case 1: f(0)=0f(0)=0. Then P(0,y)    f(f(y))+f(0)=f(y)+yf(0)    f(f(y))=f(y)P(0, y)\implies f(f(y)) + f(0)=f(y)+yf(0)\implies f(f(y))=f(y) Since the range of f:RRf:\mathbb{R}\to\mathbb{R} is R\mathbb{R}, then f(x)=x\boxed{f(x)=x}.

Case 2: f(0)=2f(0)=2. Then let g(x)=f(x)2g(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)+2yP(x):= g(g(x+y)+x+2)+f(xy)+2=x+g(x+y)+yg(x)+2y [I'll finish later]

Daniel Liu - 4 years, 3 months ago

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In case 1, you must prove that ff is surjective to make that conclusion because R\mathbb R is the codomain of ff, which is not necessarily the range.

Jubayer Nirjhor - 4 years, 3 months ago

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Oh right. I'll think about this problem some more.

Daniel Liu - 4 years, 3 months ago

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It does look like f(x)=2x f(x) = 2-x satisfies the functional equation. My guess would be f(x)=x f(x) = x or f(x)=2x 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 f(x) = x .

Patrick Corn - 4 years, 3 months ago

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Yes you are right @Patrick Corn !

f(x)=xf(x)= x or 2xx2-x\quad \forall x

Sualeh Asif - 4 years, 3 months ago

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This problem is quite a bit harder than I perceived it to be. I don't think I can find a solution.

Daniel Liu - 4 years, 3 months ago

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@Daniel Liu Daniel Liu The problem is one of those bash types on the IMO where if you keep bashing long enough you do end up on a solution! You were on the right track! Here is a solution from AOPS (by wanwan4343) which bashed in the same way! I essentially also took the same path too and ended up on the functions! (with a somewhat incomplete proof):

Let P(x,y)P(x,y) be the assertion f(x+f(x+y))+f(xy)=x+f(x+y)+yf(x)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)2f(0)=0or2P(0,0)\Rightarrow f(f(0))=0 \\P(0,f(0))\Rightarrow 2f(0)=f(0)^2\Rightarrow f(0)=0 \text{or} 2

case 1.

f(0)=2f(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)=2xRf(0)=2\Rightarrow f(2)=0 \\ P(x,1)\Rightarrow f(x+f(x+1))=x+f(x+1) \\ P(0,f(x+1)+x)\Rightarrow f(x+1)+x+2=f(x+1)+x+2(f(x+1)+x) \\ \Rightarrow f(x)=2-x\forall \in\mathbb{R}

Case 2.f(0)=0f(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)+xf(f(x)+x+1)=f(x)+x+1P(x,1)f(x+f(x1))+f(x)=x+f(x1)f(x)f(x)=f(x)P(x,x)f(x)+f(x2)=xxf(x)P(x,x)f(x)+f(x2)=x+x(x)f(x)f(x)=2xx(f(x)+f(x))f(x)=xxRP(x,0)\Rightarrow f(x+f(x))=x+f(x) \\P(x,1)\Rightarrow f(x+f(x+1))=x+1+f(x) \\ P(1,f(x+1)+x)\Rightarrow f(1+f(1+x+f(x+1)))+f(x+f(x+1))=1+f(x+1+f(x+1))+f(x+1)+x \\ \Rightarrow f(f(x)+x+1)=f(x)+x+1 \\ P(x,-1)\Rightarrow f(x+f(x-1))+f(-x)=x+f(x-1)-f(x) \\ \Rightarrow -f(x)=f(-x)\\ P(x,-x)\Rightarrow f(x)+f(-x^2)=x-xf(x)\\ P(-x,x)\Rightarrow f(-x)+f(-x^2)=-x+x(-x) \\ \Rightarrow f(x)-f(-x)=2x-x(f(x)+f(-x)) \\ \Rightarrow f(x)=x \forall x\in\mathbb{R}

hence, f(x)=2xf(x)=2-x and f(x)=xf(x)=x are solutions.

Sualeh Asif - 4 years, 3 months ago

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@Sualeh Asif Oh dang, I never thought of putting something as unusual as f(x+1)+xf(x+1)+x in the FE.

Daniel Liu - 4 years, 3 months ago

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@Daniel Liu That is what most people never thought! @Daniel Liu and that is why this is problem 5 of the IMO!

Sualeh Asif - 4 years, 3 months ago

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@Sualeh Asif Okay, so Both problems #2 and #5 were bashy. :(

Daniel Liu - 4 years, 3 months ago

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@Daniel Liu Yeah that is what I also saw! A bashing Algebra and NT problem!

However problems #1, #4 and #6 have been generally appreciated

Sualeh Asif - 4 years, 3 months ago

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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.

Gerard Renardel - 4 years, 1 month ago

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