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
Note by
Sualeh Asif
3 years, 9 months ago
No vote yet
1 vote
Easy Math Editor
*italics*or_italics_**bold**or__bold__paragraph 1
paragraph 2
[example link](https://brilliant.org)> This is a quote# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"2 \times 32^{34}a_{i-1}\frac{2}{3}\sqrt{2}\sum_{i=1}^3\sin \theta\boxed{123}Comments
Sort by:
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]
Log in to reply
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.
Log in to reply
Oh right. I'll think about this problem some more.
Log in to reply
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 \( f(x) = x \).
Log in to reply
Yes you are right @Patrick Corn !
\(f(x)= x\) or \(2-x\quad \forall x\)
Log in to reply
This problem is quite a bit harder than I perceived it to be. I don't think I can find a solution.
Log in to reply
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.
Log in to reply
Oh dang, I never thought of putting something as unusual as \(f(x+1)+x\) in the FE.
Log in to reply
That is what most people never thought! @Daniel Liu and that is why this is problem 5 of the IMO!
Log in to reply
Okay, so Both problems #2 and #5 were bashy. :(
Log in to reply
Yeah that is what I also saw! A bashing Algebra and NT problem!
However problems #1, #4 and #6 have been generally appreciated
Log in to reply
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.
Log in to reply