Let \(\mathbb{R}\) be the set of real numbers. Determine all functions \(f:\mathbb{R}\to\mathbb{R}\) 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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Sualeh Asif
3 years, 2 months ago
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Top NewestLet \(P(x,y)\) 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]
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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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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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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.
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Oh dang, I never thought of putting something as unusual as \(f(x+1)+x\) in the FE.
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That is what most people never thought! @Daniel Liu and that is why this is problem 5 of the IMO!
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Okay, so Both problems #2 and #5 were bashy. :(
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Yeah that is what I also saw! A bashing Algebra and NT problem!
However problems #1, #4 and #6 have been generally appreciated
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Yes you are right @Patrick Corn !
\(f(x)= x\) or \(2-x\quad \forall x\)
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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.
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Oh right. I'll think about this problem some more.
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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.
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