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Source: BxMO 2025 P1

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#1 h Apr 26, 2025, 11:05 AM

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Does there exist a function $f:\mathbb{R}\to \mathbb{R}$ such that $f(x^2+f(y))=f(x)^2-y$ for all $x, y\in \mathbb{R}$ ?

This post has been edited 1 time. Last edited by ErTeeEs06, Apr 26, 2025, 11:05 AM
Reason: latex mistake

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MR.1

#2 h Apr 26, 2025, 11:36 AM

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what is bxmo?

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#3 h Apr 26, 2025, 11:37 AM • 3 Y

Y by MR.1, ErTeeEs06, ATM_

Benelux ig

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Primeniyazidayi

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Primeniyazidayi

#4 h Apr 26, 2025, 11:44 AM

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My answer is no.
Let $f(a)=f(b)$ .P(x,a) and P(x,b) gives that $a=b$ ,so $f$ is injektiv.We can send $y$ anywhere,so $f$ is surjektiv.Then $f$ is bijektiv.Let $f(t)=0$ .
P(0,x) gives $f(f(x))=f(0)^2-x$ .
P(x,t) gives $f(x^2)=f(x)^2-t$
Note that $f(f(x)^2)=f(f(x))^2-t=(f(0)^2-x)^2-t=x^2-2f(0)^2x+f(0)^4-t$ .
$P(x,f(x)^2)$ gives $f(2x^2-2f(0)^2x+f(0)^4-t)=0$ .Because of the injectivity we have $2x^2-2f(0)^2x+f(0)^4=2t$ .Then $2x^2-2f(0)^2x+f(0)^4$ is constant which is impossible.

This post has been edited 3 times. Last edited by Primeniyazidayi, Apr 26, 2025, 11:48 AM

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#5 h Apr 26, 2025, 11:49 AM

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Can you explain further the proof of surjectivity?

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Primeniyazidayi

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Primeniyazidayi

#6 h Apr 26, 2025, 11:52 AM • 1 Y

Y by Rayanelba

Rayanelba wrote:

Can you explain further the proof of surjectivity?

If we fix $x$ then RHS can be any real number since we can choose $y$ as any real number.Then LHS can (and must) be any real number,which gives the result.

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#7 h Apr 26, 2025, 11:56 AM • 1 Y

Y by NicoN9

The answer is there exists no such function. If $f(a)=f(b)$ we can compare the equation with $y=a$ and $y=b$ and see $f$ is injective. Comparing $P(x, y)$ and $P(-x, y)$ gives $f(x)^2=f(-x)^2$ and injectivity implies $f$ is odd. Now compare $P(0, a)$ and $P(0, -a)$ for some real $a$ . We see $f(0)^2-a=f(f(a))=-f(f(-a))=-(f(0)^2+a)$ So $f(0)^2=-f(0)^2$ and $f(0)=0$ .

$P(1, 0)$ gives $f(1)=f(1)^2$ and by injectivity $f(1)=1$ .

Now $P(1, 1)$ gives $f(2)=0$ which contradicts injectivity. Therefore there is no function $f$ satisfying the equation.

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#8 h Apr 26, 2025, 12:18 PM • 1 Y

Y by ATM_

Another proof even if it's long :
Let $P(x,y)$ be the assertion
Claim 1 : f is injective
Proof : take a and b s.t : $f(a)=f(b)$
$P(x,a)$ and $P(x,b)$ gives a=b
Hence f is injective
$P(x,f(x)^2):f(x^2+f(f(x)^2))=0$
Taking -x we get that f is odd (with injectivity)
So : $f(f(f(0)^2))=0\implies x^2+f(f(x)^2)=f(f(0)^2)$ (*)
$P(0,y):f(f(y))=f(0)^2-y$
$P(0,0):f(f(0))=f(0)^2$
$P(0,f(0)):f^3(0)=f(0)^2-f(0)\implies f(f(0)^2)=f(0)^2-f(0)\implies f(f(0)^2-f(0))=0$
$P(x,f(0)^2-f(0)):f(x^2)=f(x)^2-f(0)^2+f(0)$
Taking $x=f(0)^2-f(0):f((f(0)^2-f(0))^2)=f(0)\implies f(0)^2=f(0)\implies f(0)=0,1$
Case 1 : $f(0)=0$
$P(x,0):f(x^2)=f(x)^2$ So f is positive for positive inputs
But : $P(0,x^2):f(f(x^2))=-x^2\implies f(f(x)^2)=-x^2$ Absurde
Case 2 : $f(0)=1$
Then : $P(0,0):f(1)=f(0)^2=1$ Absurde
Hence there exists no such function

This post has been edited 1 time. Last edited by Rayanelba, Apr 26, 2025, 12:19 PM
Reason: Typo

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NicoN9

#9 h Apr 27, 2025, 8:28 AM

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I really don't know if this is right, as I misread the problem so many times :wacko:

:wacko:

:pilot:

The answer is no. Assume for a contradiction.

claim. $f$ is bijective.
proof. $f$ is injective. Indeed, let $f(a)=f(b)$ , then we have $f(x)^2-a =f(x^2+f(a)) =f(x^2+f(b)) =f(x)^2-b$ hence $a=b$ . Also, $f$ is surjective, as if we fix $x$ , RHS moves over $\mathbb{R}$ . $\square$ .

Let $u\in \mathbb{R}$ satisfies $f(u)=0$ . Fix $y$ , and compare with $x:= -x$ , we have $f(x)^2-y=f(-x)^2-y \Longleftrightarrow f(x)^2=f(-x)^2.$ In particular, $0=f(u)^2=f(-u)^2$ , (and since $f$ is bijective, ) which implies $u=0$ .

Let $x=0$ then we get $f(f(y))=-y$ . Plugging in $y=f(y)$ , we get $f(x^2-y)=f(x)^2-f(y)$ and by $y=x^2$ , we conclude $f(x)^2=f(x^2)$ . This immediately implies $f(1)=1$ , and $f(-1)=-1$ , but putting $x=1$ , and $y=-1$ , we have $f(1-1)=0\neq 1^2+1=2.$ Contradiction.

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#10 h Apr 27, 2025, 12:00 PM

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solution

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#11 h May 2, 2025, 1:06 PM

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sol

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#12 h May 5, 2025, 4:32 PM • 1 Y

Y by IcosahedralDice

solution

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