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 \(y\).

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestLet \(P(x,y)\) 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

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

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

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

Log in to reply

@Daniel Liu and that is why this is problem 5 of the IMO!

That is what most people never thought!Log in to reply

Log in to reply

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

Log in to reply

Yes you are right @Patrick Corn !

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

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 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