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

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

for all real numbers $x$ and $y$ .

This is part of the set IMO 2015

#Algebra #Functions #FunctionalEquations #InternationalMathOlympiad(IMO) #IMO2015

Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

Markdown	Appears as
`italics` or `_italics_`	italics
`bold` or `__bold__`	bold
- bulleted - list	bulleted list
1. numbered 2. list	numbered list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1 paragraph 2	paragraph 1 paragraph 2
`[example link](https://brilliant.org)`	example link
`> This is a quote`	This is a quote
# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"	# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"

Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

Sort by:

Top Newest

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 - 4 years, 8 months ago

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 - 4 years, 8 months ago

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

Daniel Liu - 4 years, 8 months ago

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 - 4 years, 8 months ago

Yes you are right @Patrick Corn !

$f(x)= x$ or $2-x\quad \forall x$

Sualeh Asif - 4 years, 8 months ago

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, 8 months ago

@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 - 4 years, 8 months ago

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

Daniel Liu - 4 years, 8 months ago

@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, 8 months ago

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

Daniel Liu - 4 years, 8 months ago

@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, 8 months ago

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, 7 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$