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Problem 5! IMO 2015

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


for all real numbers \(x\) and \(y\).

This is part of the set IMO 2015

Note by Sualeh Asif
2 years ago

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

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

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] Daniel Liu · 2 years ago

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@Daniel Liu 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 \( f(x) = x \). Patrick Corn · 2 years ago

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@Patrick Corn This problem is quite a bit harder than I perceived it to be. I don't think I can find a solution. Daniel Liu · 2 years 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 \(P(x,y)\) be the assertion \(f(x+f(x+y))+f(xy)=x+f(x+y)+yf(x)\)

\(P(0,0)\Rightarrow f(f(0))=0 \\P(0,f(0))\Rightarrow 2f(0)=f(0)^2\Rightarrow f(0)=0 \text{or} 2\)

case 1.

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

Case 2.\(f(0)=0\)

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

hence, \(f(x)=2-x\) and \(f(x)=x\) are solutions. Sualeh Asif · 2 years ago

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@Sualeh Asif Oh dang, I never thought of putting something as unusual as \(f(x+1)+x\) in the FE. Daniel Liu · 2 years 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 · 2 years ago

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@Sualeh Asif Okay, so Both problems #2 and #5 were bashy. :( Daniel Liu · 2 years 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 · 2 years ago

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@Patrick Corn Yes you are right @Patrick Corn !

\(f(x)= x\) or \(2-x\quad \forall x\) Sualeh Asif · 2 years ago

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@Daniel Liu In case 1, you must prove that \(f\) is surjective to make that conclusion because \(\mathbb R\) is the codomain of \(f\), which is not necessarily the range. Jubayer Nirjhor · 2 years ago

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@Jubayer Nirjhor Oh right. I'll think about this problem some more. Daniel Liu · 2 years 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 · 1 year, 10 months ago

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