International FE Olympiad P5

Art of Problem Solving

AoPS Online

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

International FE Olympiad P5

G H J

Reveal topic tags

functional equation

L

G H BBookmark kLocked kLocked NReply

Source: IFEO DAY 2 P5

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

1026 posts

#1 h Feb 6, 2021, 6:05 AM • 6 Y

Y by Functional_equation, yayitsme, nikaaryu, EmilXM, megarnie, Wisphard

Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that for any reals $x,y$ , $f(x^2+f(x)+y)=xf(x)+f(x+y)$
$\textit{Proposed by gghx}$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

378 posts

EmilXM

#2 h Feb 6, 2021, 8:52 AM • 3 Y

Y by megarnie, Mango247, Mango247

Solution that I submitted.

Attachments:

Solution for P5.pdf (49kb)

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

181 posts

hyay

#3 h Feb 6, 2021, 9:20 AM • 1 Y

Y by Nartku

The only solutions are $f(x) = 0$ and $f(x) = x$ , which trivially work.

Let $P(x, y)$ be the assertion to the given equation. $P(0, x)$ gives $f(x + f(0)) = f(x)$ , so $f(f(0)) = f(-f(0)) = f(0)$ . Then, from $P(f(0), y)$ and $P(-f(0), y)$ , we have
$f(f(0)^2 + f(0) + y) = f(0)^2 + f(y) = -f(0)^2 + f(y) \Longrightarrow f(0) = 0$ Now, we work on two separate cases.

Case 1. There exists $t \neq 0, 1$ such that $f(t) = 0$ .
First, we prove the following claim.

Claim 1. If $c \in \mathbb{R}$ satisfies $f(x + c) = f(x)$ for all $x \in \mathbb{R}$ , then $f(y + 2cx) = f(y) + cf(x)$ for all $x, y \in \mathbb{R}$ . In particular, $y = 0$ gives $f(2cx) = cf(x)$ .
Proof. From $P(c, x)$ we have $f(x + c^2) = f(x + c) = f(x)$ . Then, comparing $P(x + c, y - x)$ with $P(x, y - x + 2cx)$ , we have
$f(x^2 + f(x) - x + 2cx + y) = (x + c)f(x) + f(y) = xf(x) + f(y + 2cx)$ Therefore, $f(y + 2cx) = f(y) + cf(x)$ .

Let $c = t^2 - t = t(t - 1) \neq 0$ . Then, from $P(t, x - t)$ we have $f(x + c) = f(x)$ , and thus $f(2cx) = cf(x)$ . Setting $x = 2$ gives $f(2) = 0$ (since $f(4c) = f(3c) = f(2c) = f(c) = 0$ ), and from $P(2, x - 2)$ we have $f(x + 2) = f(x)$ , and thus $f(y + 4x) = f(y) + 2f(x)$ . Furthermore,
$f(16x) = f(12x) + 2f(x) = f(8x) + 4f(x) = f(4x) + 6f(x) = 8f(x)$ but
$f(16x) = 2f(4x) = 4f(x)$ hence $\boxed{f(x) = 0}$ for all $x$ , which is a solution.

Case 2. $f(x) \neq 0$ for all $x \neq 0, 1$ .
$P(-1, 0)$ gives $f(1 + f(-1)) = 0$ , so $1 + f(-1) = 0$ or $1$ , but since $f(-1) \neq 0$ , then f(-1) = -1. Therefore, from $P(-1, x + 1)$ we have $f(x + 1) = f(x) + 1$ . Now, comparing $P(x + 1, y - x)$ with $P(x, y + x)$ , we have
$f(x^2 + f(x) + x + y) = xf(x) + x + f(x) + f(y) = xf(x) + f(2x + y)$ thus
$f(2x + y) = f(x) + f(y) + x$ In particular, $f(2x) = f(x) + x$ , so
$f(x) + 3x = f(2x) + 2x = f(4x) = f(x) + f(2x) + x = 2f(x) + 2x$ hence $\boxed{f(x) = x}$ , which is a solution.

This post has been edited 3 times. Last edited by hyay, Feb 6, 2021, 9:25 AM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

539 posts

Mr.C

#4 h Feb 6, 2021, 9:30 AM • 3 Y

Y by KaiDaMagical336, Aritra12, ywq233

Let $p(x,y)=f(x^2+f(x)-x+y)=f(y)+xf(x)$ . Assume for some reals $a,b$ where $a$ is non_zero we have :
$f(x+a)=f(x)+b$ so we have from
$T(x,y)=p(x+a,y)-p(x,y)=f(a^2+2ax+y+b-a)=xb+f(x)a+ab+f(y)$ now from $T(x,2ay)-T(y,2ax)$ we have :
$f(2ax)=c+bx+af(x)$ so rewrite $T(x,y)$ as :
$f(a^2+2ax+y+b-a)=f(2ax)-c+ab+f(y)$ or just say
$f(a^2+x+y+b-a)=f(x)-c+ab+f(y)$ from this it is easily obtained that
$f(x)+f(y)=f(x+y)+f(0)$ now we prove $f(0)=0$ . we have from $p(0,y)$ that :
$f(f(0)+y)=f(y)$ Now lets look at :
$f(4(x^2+f(x)-x)-((2x)^2+f(2x)-2x)+y)=f(y)+4(xf(x))-(2xf(2x))$ We clearly have $f(2x)=2f(x)-f(0)$ so we have
$R(x,y)=f(y+2(f(x)-x))=f(y)+2xf(0)$ now we get $f(0)=0$ with
$R(x,y)-R(x+f(0),y)$ So $f$ is additive. putting back in $p(x,y)$ we have
$Z(x)=f(x^2)+f(f(x))-f(x)=xf(x)$ so now from
$Z(x)-Z(-x)$ we have
$f(f(x))=f(x)$ so we are left with
$f(x^2)=xf(x)$ so again
$f(x^2+2x+1)=(x+1)(f(x)+f(1))=f(1)+2f(x)+f(x^2)=f(1)+2f(x)+xf(x)$ we clearly get $f(x)=f(1)x$
but also as $f(f(x))=f(x)$ we have
$f(1)^2=f(1)$ so we have $f(1)=1$ or $f(1)=0$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Functional_equation

530 posts

Functional_equation

#5 h Feb 13, 2021, 1:08 PM

Y by

This is IFEO SL A4

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

11102 posts

#6 h Aug 31, 2021, 4:27 PM • 1 Y

Y by hakN

Remark

Let $P(x,y)$ denote the assertion $f(x^2+f(x)+y)=xf(x)+f(x+y)$ .

Claim 1: $f(0)=0$
$P(0,x)\Rightarrow f(x+f(0))=f(x)\Rightarrow f(f(0))=f(0)=f(-f(0))$
$P(f(0),x)-P(-f(0),x)\Rightarrow f(0)=0$

Claim 2: If $f(x+a)=f(x)$ for all $x$ , where $a$ is some nonzero constant, then $f(x)=0$ .
$P(x+a,y-x^2-f(x))-P(x,y-x^2-f(x))\Rightarrow f(2xa+y)=f(y)+af(x)\Rightarrow f(2xa)=af(x)$
Setting $x=\frac12$ and using the fact that $f(a)=0$ , we have $f\left(\frac12\right)=0$ . Then $f(2xa+y)=f(y)+f(2xa)$ , so $f$ is additive.
$P(x+y,0)-P(x,0)-P(y,0)\Rightarrow f(2xy)=xf(y)+yf(x)$ , setting $y=\frac12$ we obtain $\boxed{f(x)=0}$ , which fits.

Assume that $f$ is not identically zero, so $f(x+a)=f(x)$ implies $a=0$ .
Assume $f(k)=0$ for some $k$ .
$P(k,x-k)\Rightarrow f(x+k^2-k)=f(x)\Rightarrow k\in\{0,1\}$
$P(-1,0)\Rightarrow f(f(-1)+1)=0\Rightarrow f(-1)\in\{-1,0\}$

If $f(-1)=0$ , then $P(-1,x+1)\Rightarrow f(x+2)=f(x)$ , contradiction.

Now assume $f(-1)=-1$ :
$P(-1,x+1)\Rightarrow f(x+1)=f(x)+1$
$P(x+1,y-x^2-f(x))-P(x,y-x^2-f(x))\Rightarrow f(2x+y)=f(x)+x+f(y)$ , setting $y=-x$ gives the solution $\boxed{f(x)=x}$ , which is a solution.

This post has been edited 1 time. Last edited by jasperE3, Aug 31, 2021, 4:47 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

5538 posts

#8 h Jan 7, 2022, 4:58 PM

Y by

The only solutions are $f\equiv 0$ and $f(x) = x$ . These work. Now we show they are the only ones.

Assume $f$ isn't constant.

Claim: If $f$ is additive, then $f(x) = x$ for all reals $x$ .
Proof: Suppose $f$ was additive. Then the equation becomes $f(x^2) + f(f(x)) + f(y) = xf(x) + f(x) + f(y)$ , so $f(x^2) + f(f(x)) = xf(x) + f(x)$ Setting $x = 1$ gives $f(f(1)) = f(1)$ .

Setting $x \to x + 1$ gives $f((x+1)^2) + f(f(x + 1)) = (x+1) f(x+1) + f(x+1)$ .

The LHS becomes $f(x^2 + 2x + 1) + f(f(x) + f(1)) = f(x^2) + 2f(x) + 2f(1) + f(f(x))$ . SInce $f(f(x)) + f(x^2) = xf(x) + f(x)$ , this equals $xf(x) + 3f(x) + 2f(1)$ .

The RHS becomes $(x+2) (f(x) + f(1)) = xf(x) + 2f(x) + xf(1) + 2f(1)$ .

Setting them equal and subtracting both sides by $xf(x) + 2f(x) + 2f(1)$ , we get $f(x) = xf(1)$ . Setting $x = f(1)$ gives $f(1) = f(1)^2$ , so $f(1) \in \{0, 1\}$ . If $f(1) = 0$ , then $f$ is constant, absurd. Thus, $f(x) = x$ .

Claim: $f(0) = 0$
Proof: Suppose otherwise.

$P(0, x)$ gives $f(x) = f(x+c)$ for all $x$ , where $c = f(0)\ne 0$ .

$P(x+c, y): f((x+c)^2 + f(x) + y) = (x+c)f(x) + f(x+y),$ so $f(x^2 + f(x) + y + 2xc + c^2) = xf(x) + f(x+y) + cf(x)$ .

Hence $f(x^2 + f(x) + y + 2xc + c^2) - f(x^2 + f(x) + y) = cf(x)$ .

Since $x^2 + f(x) + y$ is surjective over reals, we have $f(y + 2xc + c^2) - f(y) = cf(x)$ for any reals $x,y$ .

Comparing $x$ and $x + 1$ here gives that $cf(x) = cf(x+1)$ , so $f(x) = f(x + 1)$ . Thus, $f(1) = c$ .

$P(1, y): f(y + f(1) + 1) = f(y + 1) + f(1)$ . Thus, $f(y + c) = f(y) + c$ , absurd since $f(y + c) = f(y)$ and $c\ne 0$ . $\square$

Claim: $f$ is not periodic
Proof: Suppose otherwise and $f$ was periodic with period $d\ne 0$ .

Similarly to above, $P(x+d, y) - P(x,y)$ gives $f(x^2 + f(x) + y + 2xd + d^2) - f(x^2 + f(x) + y) = df(x)$ , so $f(y + 2xd + d^2) - f(y) = d f(x)$ for any reals $x,y$ . Now, setting $x = 0$ gives that $f(y + d^2) = f(y)$ , so now we have $f(y + 2xd) - f(y) = d f(x)$ Setting $y = 0$ gives $f(2xd) = df(x)$ . Hence $f(y) + f(2xd) = f(y + 2xd)$ , implying $f$ is additive, so $f(x) = x$ , implying $f$ is not periodic, absurd. $\square$

Claim: $f$ is trijective at $0$ (if $f(x) = 0$ , then $x = 0$ or $x = 1$ ).
Proof: If $f(x) = 0$ , then $f(x^2 + y) = f(x + y)$ , so $f$ is periodic with period $x^2 - x$ , implying $x\in \{0,1\}$ . $\square$

Claim: $f(1) = 1$
Proof: $P(-1,0): f(1 + f(-1)) = 0$ . Since $f(-1) \ne 0$ , we must have $f(-1) = -1$ .

$P(-1,y): f(y) = f(y - 1) + 1$ . But then setting $y = 1$ gives $f(1) = f(0) + 1 = 1$ , absurd. $\square$

Therefore, $f$ is injective at $0$ .

$P(1, y): f(y + f(1) + 1) = f(y + 1) + f(1)$

Setting $y = -f(1)$ gives $f(1 - f(1)) = 0$ , so $f(1) = 1$ .

$P(1, y - 1): f(y + 1) = f(y) + 1$ .

$P(-1,0): f(1 + f(-1)) = 0$ , so $f(-1) = -1$ .

$P(x,0): f(x^2 + f(x)) = (x+1) f(x)$ .

$P(x+1, y): f((x+1)^2 + f(x) + 1 + y) = (x+1) (f(x) + 1) + f(x+y) + 1 = xf(x) + f(x) + x + f(x+y) + 2$ .

The LHS becomes $f(x^2 + f(x) + y + 2x + 2) = f(x^2 + f(x) + y + 2x) + 2$ .

Now subtracting by $2$ both sides gives that $f(x^2 + f(x) + y + 2x) = xf(x) + f(x) + x + f(x+y)$ . Comparing with original equation gives that $f(x^2 + f(x) + y + 2x) - f(x^2 + f(x) + y) = x + f(x)$ .

Since $x^2 + f(x) + y$ is surjective over reals, we have $f(y + 2x) - f(y) = x + f(x)$ . Now setting $y = 0$ gives that $f(2x) = x + f(x)$ . Thus, $f(y + 2x) = f(y) + f(2x)$ , meaning $f$ is additive, and therefore $f(x) = x$ .

This post has been edited 2 times. Last edited by megarnie, Aug 14, 2024, 10:33 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

355 posts

#9 h Oct 1, 2023, 7:28 AM

Y by

Storage

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a