weird FE on R

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

weird FE on R

G H J

Reveal topic tags

Functional Equations

L

G H BBookmark kLocked kLocked NReply

Source: probably own

The post below has been deleted. Click to close.

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

60 posts

frac

#1 h Jan 4, 2025, 12:15 PM • 1 Y

Y by NicoN9

Find all functions $f:\mathbb{R} \rightarrow \mathbb{R}$ such that
$f(x+y)^2=xf(x+f(y))+yf(f(y))+f(xy)$ for all $x,y\in \mathbb{R}$ .

This post has been edited 4 times. Last edited by frac, Mar 23, 2025, 7:22 AM

Z K Y

The post below has been deleted. Click to close.

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

666 posts

#2 h Jan 4, 2025, 1:24 PM

Y by

$f(x+y)^2=xf(x+f(y))+yf(f(y))+f(xy)$ Answers are $f\equiv 0,f(x)=x$ and $f(x)=x+1$ which indeed fit. Let $P(x,y)$ be the assertion. Plugging $x=y=0$ gives $f(0)=0$ or $f(0)=1$ . We split into two cases.

Case $I$ : If $f(0)=0$ , then $P(x,0)$ yields $f(x)^2=xf(x)\iff f(x)(f(x)-x)=0$ so $f(x)\in \{0,x\}$ . Suppose that there exists $f(a)=0$ and $a\neq 0$ . Pick $y=a$ which implies $f(x+a)^2=xf(x)+f(xa)$ . Subsituting $x=a$ gives $f(2a)^2=f(a^2)=af(a)=0$ thus, $f(a)=0$ implies $f(2^ka)=0$ . If there exists $f(b)=b$ where $b\neq 0$ , then $x=b$ yields $f(a+b)^2=b^2+f(ab)$ . Left hand side is one of $\{0,a^2+2ab+b^2\}$ and right hand side is one of $\{b^2,b^2+ab\}$ . Hence $a+b=0$ or $a+2b=0$ . We can pick $a$ such that $a\neq -b,-2b$ since there $f(2^ka)=0$ . Thus, there cannot be $f(a)=0$ and $f(b)=b$ where $a\neq 0\neq b$ at the same time so we get the solutions $f\equiv 0$ and $f(x)=x$ .

$f(x+y)^2=xf(x+f(y))+yf(f(y))+f(xy)$ Case $II$ : If $f(0)=1$ . $P(-1,0)$ gives $f(-1)^2=-1+1=0$ thus, $f(-1)=0$ .
$yf(y+f(x))+xf(f(x))=f(x+y)^2-f(xy)=xf(x+f(y))+yf(f(y))$ Plug $y=0$ in order to see that $xf(f(x))=xf(x+1)$ . Since $f(0)=1$ , both the cases $x=0$ and $x\neq 0$ yield $f(f(x))=f(x+1)$ .
Claim: $f$ is injective.
Proof: Assume that $f(t)=0$ . Pick $x=0,y=t$ which yields $0=t+1$ iff $t=-1$ so $f$ is injective at $0$ .
Suppose that $f(a)=-1$ . Subsituting $x=1,y=a$ implies $f(1+a)^2=1+f(a)=0$ hence $f(a+1)^2=0$ iff $a=-2$ thus, $f$ is injective at $-1$ .
Now assume $f(a)=c=f(b)$ and $c\neq 0,-1$ . Plug $x=1$ in order to get
$f(f(y))^2=f(y+1)^2=f(1+f(y))+yf(f(y))+f(y)$ $y=a$ and $y=b$ give $af(c)=bf(c)$ and since $c\neq -1$ , we have $f(c)\neq 0$ which implies $a=b$ . $\square$
Since $f$ is injective and $f(f(x))=f(x+1)$ , we have $f(x)=x+1$ as desired. $\blacksquare$

Z K Y

The post below has been deleted. Click to close.

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

optimusprime154

17 posts

optimusprime154

#3 h Jan 4, 2025, 2:58 PM • 1 Y

Y by SorPEEK

let $P(x,y)$ be the assertion.
$P(0,0)$ yields $f(0) = 1$ or $f(0) = 0$
case 1: $f(0) = 1$ :
$P(0,x)$ gives $f(x)^2 = xf(f(x)) + 1$ (1)
$P(x,0)$ gives $f(x)^2 = xf(x+1) + 1$ (2)
combining those two we get $f(x+1) = f(f(x))$ (3)
suppose there exists a number c such that $f(f(c)) = 0$ from the above statements we obtain $f(c+1) = 0$ and $f(c) = 1$ or $f(c) = -1$
subcase 1.1:
$f(c) = 1$
take $P(1, c)$
this gives: $f(c+1)^2 = f(2) + 1$ or $f(2) = -1$ this means from (2) that $f(1) = 0 => f(f(1)) = 1$ using $P(0,1)$ gives us $0 = 1 + 1$ which is clearly a contradiction.
subcase 1.2:
$f(c) = -1$
we get $f(-1) = 0$ and $f(f(-1) = 0$ comparing $P(x, -1)$ and (1) we get $-f(x) + 2$ = $f(-x)$ meaning $f(1) = 2$
comparing $P(x,1)$ with $x+1$ in (2) we get $f(x+2) = f(x) + 2$ (4)
now suppose there exist 2 non equal numbers $a$ and $b$ such that $f(a) = f(b)$ if they are both equal to 0 we can use (4) to switch them while keeping $f(a+2) = f(b+2)$
now we have our desired numbers such that they dont map to zero.
compare $P(1,a)$ with $P(1,b)$ to get $f(a) * a = f(b) * b$ meaning $a = b$ and so injectivity. using injectivity in (3) we get $f(x) = x+1$ which is indeed a solution.
case 2: $f(0) = 0$
$P(x, 0)$ gives $f(x) = 0$ or $f(x) = x$ suppose there exist 2 numbers $u$ and $v$ such that $-u$ does not equal $v$ and $f(u) = 0$ and $f(v) = v$
consider $P(u,v)$ we get $f(u+v)^2 = uf(u+v) + v^2 + f(uv)$ now its obvious $f(u+v)$ cant map to zero so it has to map to $u+v$ and $f(uv) = uv$ but
$P(v,u)$ tells us $f(u+v)^2 = v^2 + f(uv)$ meaning $u^2 + uv = 0$ which is not true unless $u = -v$ but we can always choose these numbers such that the equality doesnt hold.
so our answers are $f(x) = x, f(x) = 0. f(x) = x +1$

Z K Y

The post below has been deleted. Click to close.

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

60 posts

frac

#4 h Mar 23, 2025, 6:40 AM

Y by

bin_sherlo wrote:

...Subsituting $x=a$ gives $f(2a)^2=f(a^2)=af(a)=0$ ...

Could you explain why $f(a^2)$ implies $af(a)$ ? (Sorry for 3 months bump btw :oops_sign:

:oops_sign:

)

Z K Y

The post below has been deleted. Click to close.

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

83 posts

NicoN9

#5 h Mar 23, 2025, 7:48 AM

Y by

I recently practice FE, but this one is nice! Good problem for beginners, imo.

The answer is $f(x)\equiv 0, x, x+1$ . These works, since $f\equiv 0$ is trivial, and that $\begin{align*} (x+y)^2&=x^2+xy+y^2+xy, \\ (x+y+1)^2&=x^2+xy+2x+y^2+2y+xy+1. \end{align*}$ We prove that these are only the solution. Let $P(x, y)$ denote the assertion.
First, we easily see that $f(0)\in \{0, 1\}$ , by $P(0, 0)$ .

Now we casework wrt the value of $f(0)$ .

case1: $f(0)=0$ .
We prove that $f(x)=0$ , or $f(x) = x$ . First, from $P(x, 0)$ , we have $f(x)^2=xf(x)\Longleftrightarrow f(x)\in \{0, x\}.$ Let $u\neq 0$ be a real number such that $f(u)=0$ .It is suffice to show that $f(x)=0$ for all $x\in \mathbb{R}$ . Suppose $f(x)=x$ for some $x\neq 0$ . From $P(x, u)$ , we have $f(x+u)^2=x^2+f(xu).$ We have four cases now, namely $f(x+u)=0$ or $x+u$ , and $f(xu)=0$ or $xu$ . Checking all of these cases, we see that this is only possible if $x=-u$ . Then pick a real number $v\neq u$ , $v\neq 0$ and $f(v)=0$ . Taking $P(x, v)$ to get $f(-u)=0$ , and we are done.

case2: $f(0)=1$ .
This one is easier, probably. In this case, we prove that $f(x)=x+1$ . Compare $P(0, x)$ and $P(x, 0)$ , we have $\begin{align*} xf(x+1)+1=f(x)^2&=xf(f(x))+1 \end{align*}$ So now $f(x+1)=f(f(x))$ .

claim. $f$ is injective.
proof. By $P(0, -1)$ and $P(1, -1)$ , we see that $f(-1)=0$ and $f(1)=2$ . Let $t$ be real number with $f(t)=0$ . Then combine $P(0, t)$ and $P(1, t)$ , we get $t = -1$ , only. Similarly, If we have $f(s)=-1$ for some $s$ , then $P(0, s)$ to get $s=-2$ , only.
Let $a, b$ be reals such that $f(a)=f(b)$ , and both not $-1$ , or $-2$ . Now by $P(0, x)$ , $af(f(a))+f(0)=f(a)^2=f(b)^2=bf(f(b))+f(0)$ hence $a=b$ , as desired. (We used the fact that $f(f(a)) \neq 0$ ).

So now $f(x+1)=f(f(x))$ implies $f(x)=x+1$ .

As these cover all the cases, we are done.

P.S. I think it is fixed now.

This post has been edited 2 times. Last edited by NicoN9, Mar 23, 2025, 9:28 AM
Reason: claim was wrong!

Z K Y

The post below has been deleted. Click to close.

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

60 posts

frac

#6 h Mar 23, 2025, 8:31 AM

Y by

NicoN9 wrote:

claim. $f$ is injective.
proof. Let $a, b$ be reals such that $f(a)=f(b)$ . Now by $P(0, x)$ ,
$af(f(a))+f(0)=f(a)^2=f(b)^2=bf(f(b))+f(0)$ hence $a=b$ , as desired.

@above
Umm, I think you need a little more work to prove this claim.
More precisely, you proved this claim when $f(f(a))=f(f(b)) \neq 0$ and what if $f(f(a))=f(f(b))=0$ ?

Z K Y

The post below has been deleted. Click to close.

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

83 posts

NicoN9

#7 h Mar 23, 2025, 9:08 AM

Y by

frac wrote:

NicoN9 wrote:

claim. $f$ is injective.
proof. Let $a, b$ be reals such that $f(a)=f(b)$ . Now by $P(0, x)$ ,
$af(f(a))+f(0)=f(a)^2=f(b)^2=bf(f(b))+f(0)$ hence $a=b$ , as desired.

@above
Umm, I think you need a little more work to prove this claim.
More precisely, you proved this claim when $f(f(a))=f(f(b)) \neq 0$ and what if $f(f(a))=f(f(b))=0$ ?

Oh yeah thanks. I should do this on case2, since checking the injectivity knowing that $f\equiv 0$ is a solution is crazy

Also, I actually used this only in case2.
I think it is fixed now.

This post has been edited 1 time. Last edited by NicoN9, Mar 23, 2025, 9:29 AM

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a