Functional equations in IMO TST

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sheripqr

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sheripqr

#1 h Sep 14, 2015, 8:06 PM • 6 Y

Y by langkhach11112, Adventure10, ATGY, ItsBesi, D_S, Tastymooncake2

Find all functions $f: \mathbb R \to \mathbb R$ such that $f(f(x)+y)=f(x^2-y)+4f(x)y$ for all $x,y \in \mathbb R$

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Reason: Typo

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efang

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efang

#2 h Sep 14, 2015, 11:14 PM • 3 Y

Y by Delray, Adventure10, Mango247

$f(f(x)+y) = f(x^2-y) + 4f(x)y$

Plug in $y=x=0$ to get $f(f(0)) = f(0)$

Plug in $x=0, y=-f(0)$ to get $f(0) = f(f(0)) - 4f(0)^2$ . Thus $4f(0)^2 = 0$ or $f(0) = 0$

Plug in $y=-f(x)$ we have $f(x^2+f(x)) - 4f(x)^2 = 0$ or $f(x^2+f(x)) = 4f(x)^2$ .

Plug in $y=x^2$ to get $f(f(x)+x^2) = 4x^2f(x)$

Thus $4x^2f(x) = 4f(x)^2$ . So either $f(x) \equiv 0$ or $4f(x) = 4x^2 \implies f(x) \equiv x^2$

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v_Enhance

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v_Enhance

#3 h Sep 15, 2015, 12:30 AM • 5 Y

Y by adik7, Jc426, HamstPan38825, Adventure10, Mango247

Plug in $y = \frac12(x^2-f(x))$ to get directly that $f(x) \in \{0,x^2\}$ for every $x$ .

If $f(1)=1$ , then $f(1+y)=f(1-y)+4y$ for every $y$ ; then clearly RHS is nonzero for each $y \neq 0, -1$ which gives $f(x) \equiv x$ .
If $f(1) = 0$ , then $f(y) = f(1-y)$ for every $y$ , so $f(y) = 0$ unless $y^2 = (1-y)^2$ ; but take $x=2$ to get $f(y) = f(4-y)$ ; conclude then that $f(x) \equiv 0$ .

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pco

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pco

#4 h Sep 15, 2015, 5:03 AM • 5 Y

Y by torquoiseworld, Tawan, Kingsbane2139, Adventure10, Mango247

efang wrote:

... Thus $4x^2f(x) = 4f(x)^2$ . So either $f(x) \equiv 0$ or $4f(x) = 4x^2 \implies f(x) \equiv x^2$

Not exactly.
$4x^2f(x) = 4f(x)^2$ $\forall x$ implies that " $\forall x$ , either $f(x)=0$ , either $f(x)=x$ "
which is different from what you wrote : "either $f(x)=0$ $\forall x$ , either $f(x)=x$ $\forall x$ "
(for example $f(x)=x\times 1_{\mathbb Q}(x)$ matches the equation $4x^2f(x) = 4f(x)^2$ $\forall x$ ).

It remains a little job to do ...

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julianf

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julianf

#5 h Feb 23, 2016, 4:11 AM • 2 Y

Y by Adventure10, Mango247

You also have to demostrate that is imposible to have both $f(a)=0$ and $f(b)=b^2$ for $a \neq b$ and $a,b \neq 0$
It is not difficult... Only putting that in the original func.eq and you'll find a contradiction:

$P(b,a)$ is equivalent to $f(b)=f(a^2-b)$ which is $b^2= (a^2-b)^2$ (because in the other case b^2=0)

$b^2=(a^2-b)^2=a^4-2a^2b+b^2 \rightarrow a^2(a^2-2b)=0$ but this means that for every $b$ such $f(b)=b^2$ and $a$ such that $f(a)=0$ we have $b^2=2a$ ... Contradiction

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Zorger74

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Zorger74

#6 h Nov 25, 2020, 3:20 AM

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Solution

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Ucchash

181 posts

Ucchash

#7 h Nov 25, 2020, 11:32 AM

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$P(x,y)$ be the assertion,
$P(0,0)$ :
$f(f(0))=f(0)$
Let, $f(0)=c$
$P(0,-c)$ :
$f(0)=f(c)+4f(c)(-c)$
Or, $-4c^2=0$
$c=f(0)=0$
$P(x,0)$ :
$f(f(x))=f(x^2)$

$P(x,x^2-f(x))$ :
$f(x^2)=f(f(x))+4f(x)(x^2-f(x))$
Or, $f(x)(x^2-f(x))=0$
From this, $f(x)=0$ and $f(x)=x^2$ which are solution

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Zorger74

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Zorger74

#8 h Nov 25, 2020, 1:22 PM

Y by

@above you missed the pointwise trap

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Ucchash

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Ucchash

#9 h Nov 25, 2020, 1:55 PM

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Zorger74 wrote:

@above you missed the pointwise trap

Can you please help me to complete the proof? actually i am not so much skilled in FE, so i missed that :sad

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rafaello

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rafaello

#10 h Nov 25, 2020, 8:26 PM

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Let $P(x,y)$ be the assertion of $f(f(x)+y)=f(x^2-y)+4f(x)y$ .
$P(x,0)\implies f(f(x))=f(x^2)$ .
$P(x,-f(x))\implies f(0)=f(x^2+f(x))-4f(x)^2$ .
Thus, $f(f(0))=f(0)$ and $f(0)=f(f(0))+-4f(0)^2$ , which means that $f(0)=0$ .
$P(0,y)\implies f(y)=f(-y)$ .
$P(x,x^2)\implies f(x^2+f(x))=4f(x)x^2$ .
Thus, we have that $4f(x)x^2-4f(x)^2=0 \implies f(x)=0 \text{ or } f(x)=x^2$ .

If there are $f(a)=a^2$ , $f(b)=0$ .
$P(b,b^2-a)\implies f(b^2-a)=a^2$ , thus $a=0$ , $b=0$ or $b^2=2a$ .
If $f(c)=c^2$ , where $a\neq c$ , thus we $a=c$ , contradiction.
Only option from here is that $f(x)=0$ for all $x$ , not $a$ .
$P(0,a)\implies a^2=f(a)=0\implies a=0$ . Thus from here we have that $f(x)=0$ .

$f(x)=x^2$ means that:
$\mathcal{RHS}=f(f(x)+y)=f(x^2+y)=(x^2+y)^2=x^4+y^2+2x^2y$ and
$\mathcal{LHS}=f(x^2-y)+4f(x)y =(x^2-y)^2+4x^2y=x^4+y^2+2x^2y$ , thus $\mathcal{RHS}=\mathcal{LHS}$ .

Answer. $\boxed{f(x)=0\,\forall x\in \mathbb R \text{ and }f(x)=x^2\,\forall x\in \mathbb R}$

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EulersTurban

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EulersTurban

#11 h Nov 25, 2020, 8:33 PM

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The solutions are $f(x)=0$ and $f(x)=x^2$ , for all $x$ which are in $\mathbb{R}$ .

We plug in $y=x^2$ , to get that $f(f(x)+x^2)=f(0)+4x^2f(x)$ .
Next we plug in $y=-f(x)$ , to get that $f(0)+4f(x)^2=f(x^2+f(x))$ .
From these relations we get that $f(x)^2=x^2f(x)$ .

Which brings us to our first case scenario, so that for the vast majority possible values for $x$ we have that $f(x)=0$ , but there exists some $k$ such that $f(k)=k^2$ .
We plug in $x=k$ , but we plug in $y$ so that $f(k^2-y)=0$ and so that $f(f(k)+y)=0$ , to get that $4k^2y=0$ , but this isn't possible for $k \neq 0$ and for $y \neq 0$ , so that means that $k = 0$ .

The next case scenario is that for the vast majority possible values of $x$ satisfy $f(x)=x^2$ , but there exists $k \neq 0$ such that $f(k) = 0$ .
By carefully plugging values we get that $k^2(k^2-2y)=0$ , since $k \neq 0$ (by our assumption) we must have that $k^2=2y$ , but just set $y < 0$ , thus we get a massive contradiction.

Thus that means that the solutions aren't piecewise functions which brings us to our already stated solutions.

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megarnie

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megarnie

#12 h Apr 9, 2022, 6:49 PM

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Solved with Ritwin.

$P\left(x,\frac{f(x)-x^2}{2}\right): 4f(x)\frac{f(x)-x^2}{2}=0$ .

Thus, either $f(x)=0$ or $f(x)=x^2$ for each $x$ .

Suppose FTSOC there exist $a,b\ne 0$ with $f(a)=a$ and $f(b)=b^2$ .

$P(a,-b): f(a^2-b)=b^2$ .

Since $b\ne 0$ , we have $f(a^2-b)=(a^2-b)^2$ .

Thus, either $a^2-b=b$ or $a^2-b=-b$ . The latter gives $a=0$ , a contradiction. So the former must
hold true, which implies $a^2=2b$ .

So $a^2=2b$ , if $f(a)=0$ and $f(b)=b^2$ .

$P(a,b): f(3b)=f(-b)$ .

Clearly $(3b)^2\ne (-b)^2$ , so $f(3b)=f(-b)=0$ .

Since $f(-b)=0$ , we have $(-b)^2=2b\implies b=2$ .

Since $f(3b)=0$ , we have $(3b)^2=9b^2=2b\implies b=\frac{2}{9}$ .

These contradict each other.

Thus, $\boxed{f\equiv 0}$ or $\boxed{f(x)=x^2}$ , which both fit.

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Taco12

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Taco12

#13 h Jan 14, 2023, 1:26 AM

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Let $P(x,y)$ be the assertion.

$P\left(x, \frac{1}{2}f(x)-\frac{1}{2}x^2\right)\rightarrow f(x)=0,x^2.$ Now pointwise. Suppose there exists $a,b\neq 0$ so that we have $f(a)=0$ and $f(b)=b^2$ . $P(a,-b)$ gives a contradiction, so $f(x)=0$ and $f(x)=x^2$ , which work, are the only solutions.

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HamstPan38825

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HamstPan38825

#14 h Mar 25, 2023, 2:08 AM

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The solutions are $f(x) = 0$ and $f(x) = x^2$ . Using the cancellation trick with $y = \frac{f(x) - x^2}2$ , $2f(x)(f(x) - x^2) = 0,$ so $f(x) = 0$ or $f(x) = x^2$ for each $x$ . On the other hand, if $f(a) = 0$ , $f(b) = b^2$ , then $f(a^2-b) = f(b) = b^2 \neq 0,$ contradiction.

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Mogmog8

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Mogmog8

#15 h Apr 15, 2023, 4:53 AM • 1 Y

Y by centslordm

We claim the solutions are $f(x)=x^2$ and $f(x)=0$ , which both work. Let $P(x,y)$ denote the given assertion. From $P\left(x,\frac{f(x)-x^2}{2}\right)$ we see $4f(x)\left(\frac{f(x)-x^2}{2}\right)=0$ Hence, $f(x)\in\{0,x^2\}$ for each $x\in\mathbb{R}$ . Therefore, $f(0)=0$ so $P(0,y)$ yields that $f$ is even.

Suppose FTSOC that $f(a)=a^2$ and $f(b)=0$ for some some nonzero $a$ and $b$ such that $a\neq b$ . From $P(b,-a)$ we see $f(b^2-a)=f(f(b)+a)+4f(b)(-a)=a^2$ Since $a$ is nonzero, we have $(b^2-a)^2=a^2$ so $b^4=2ab^2$ so $b^2=2a$ . From $P(b,a)$ we find $f(b^2+a)=f(f(b)-a)+4f(b)a\implies f\left(\frac{3}{2}b\right)=f(-a)=f\left(\frac{1}{2}b^2\right)$ This means either $(3b/2)^2=(b^2/2)^2$ or $f(3b/2)=f(b/2)=0$ . In the former case, $b=3$ or $b=-3$ so $a=9/2$ . But from $P(b,b)$ we have $f(b^2+b)=f(-b)$ and we note both functions must be zero as the squares of the inputs cannot be equal, so $f(-b)=f(b^2+b)=0$ . Hence, $(b^2+b)^2=2a$ as $(a,b^2-b)$ is another pair of integers that have outputs $a^2$ and $0$ . Notice none of $b=-3,3$ satisfy $b^2+b=\sqrt{9/2}$ so we have a contradiction.

Hence, $f(x)=x^2$ for all $x$ or $f(x)=0$ for all $x$ . $\square$

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reni_wee

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reni_wee

#40 h Jun 3, 2024, 6:18 PM

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Plug in x=y=0,
$f(f(0)+0)=f(0^2-0)+4f(0)\cdot 0\implies f(f(0))=f(0)$ Substitute $x=0,y=-f(0),$
$f(f(0)-f(0))=f(0^2+f(0))-4f(0)^2 \implies 4f(0)^2=0 \implies f(0)=0$ Now substitute $y=-f(x),$
$f(f(x)-f(x))=f(x^2+f(x))-4f(x)^2 \implies f(x^2+f(x))=4f(x)^2$ Substituting $y=x^2$ gives us,
$f(f(x)+x^2)=f(x^2-x^2)+4x^2f(x) \implies f(x^2+f(x))=4x^2f(x)$ Hence, $4f(x)^2=4x^2f(x)\implies f(x)=0$ or $f(x)=x^2$ for all $x\in\mathbb{R}.$

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Markas

105 posts

Markas

#41 h Aug 20, 2024, 8:07 PM

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Let x = 0. We get $f(y) = f(f(0) - y) + 4f(0)y$ . Let y = 0. We get $f(f(x)) = f(x^2)$ . Let $y = f(x)$ . We have $f(f(x) + x^2) = f(0) + 4f(x)^2$ $\Rightarrow$ $f(0) = f(f(x) + x^2) - 4f(x)^2$ and plugging in x = 0 we get $f(0) = f(f(0)) - 4f(0)^2$ and since $f(0) = f(f(0))$ , we have $4f(0)^2 = 0$ $\Rightarrow$ $f(0) = 0$ . When x = 0 we got $f(y) = f(f(0) - y) + 4f(0)y$ and since $f(0) = 0$ , we get $f(y) = f(-y)$ . Let $y = - x^2$ . We get $f(0) = f(f(x) + x^2) - 4x^2f(x)$ and plugging in what we know we get $0 = 4f(x)^2 - 4x^2f(x)$ $\Rightarrow$ $f(x)^2 = x^2f(x)$ $\Rightarrow$ $f(x) = 0$ or $f(x) = x^2$ . Here we are at the pointwise trap. Let there be $f(a) = a^2$ , $f(b) = 0$ . Let x = b and $y = a - b^2$ $\Rightarrow$ we get $f(a) = f(f(b) + b^2 - a) + 4f(b)(a - b^2)$ $\Rightarrow$ $a^2 = f(b^2 - a)$ $\Rightarrow$ a = 0, b = 0 or $b^2 = 2a$ . If $f(c) = c^2$ and $a\neq c$ , we get $a = c$ - impossible $\Rightarrow$ we have $f(x) = 0$ for all x. Let x = 0, y = a, we have $a^2 = f(a) = 0$ $\Rightarrow$ a = 0 $\Rightarrow$ we have that $f(x)=0$ . Now we have to check if they work. For $f(x) = 0$ its obvious. For $f(x) = x^2$ we have $x^4 + y^2 + 2x^2y = (x^2)^2 + y^2 - 2x^2y + 4x^2y$ , which is obviously true $\Rightarrow$ $f(x) = 0$ for all $x \in R$ and $f(x) = x^2$ for all $x \in R$ . We are ready.

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cursed_tangent1434

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cursed_tangent1434

#42 h Aug 31, 2024, 3:50 PM

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The answer is $f(x) = x^2$ for all $x \in \mathbb{R}$ or $f(x)=0$ for all $x \in \mathbb{R}$ . It’s easy to see that these functions satisfy the given equation.
First we plug in $x=y=0$ to obtain $f(0)=f(f(0))$ Next, we plug in $x=0$ and $y=f(0)$ to obtain
$f(f(0))=f(0) +4f(0)^2$ But, this means that $4f(0)^2=0$ and in turn that $f(0)=0$
Next, we let $y=f(x)$ to obtain $f(x^2+f(x))=f(0)+4f(x)^2=4f(x)^2$ Then, we let $y=-x^2$ from which we obtain
$f(0)=f(f(x)+x^2)-4f(x)x^2 \implies f(f(x)+x^2) = 4f(x)x^2$ But, then notice that this means for $f(x)\neq 0$ $4f(x)^2 = 4f(x)x^2 \implies (x)=x^2$ Thus, $f(x)=0$ or $f(x)=0$ or $f(x)=x^2$ . Now, assume that it is possible for both these to happen simultaneously. Setting $x=0$ , we see that $f(y)=f(-y)$ Then assuming $f(a)=0$ and $f(b)=b^2$ ,
$\begin{align*} f(a^2+b)= f(f(a)-b) + 4f(a)b\\ f(a^2+b)=f(-b)\\ f(a^2+b)=b^2 \end{align*}$ which is clearly impossible irrespective of which of the two possible values is taken by $f(a^2+b)$ (requires $b=0$ ), and thus contradiction.
Thus, we have that all the functions which satisfy the required equation must be of the claimed type.

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N3bula

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N3bula

#43 h Sep 18, 2024, 9:52 AM

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Let $P(x, y)$ denote the assertion in the question.
$P(0, 0)$
$f(0)=f(f(0))$ $P(0, f(0))$
$4f(0)^2=0$ $P(0, y)$
$f(y^2+f(y))=4f(x)^2$ $P(y, y^2)$
$f((y)+y^2)=4f(x)x^2$ thus
$f(x)^2=f(x)x^2$ So we get $f(x)\in \{0, x^2\}$ , now suppose there exists a value $u$ such that $u\neq 0$ and $f(u)=0$ .
$P(u, y)$
$f(u^2+y)=f(-y)$ As $f(x)=f(-x)$ we can wlog that $y$ is always positive above, thus as $(u^2+y)^2\neq y^2$ for any positive $y$ we get that $f(-y)=0$ for all positive $y$ and thus that
$f(x)=0$ for all $x$ . Thus the solutions are $f(x)=0$ for all $x$ or $f(x)=x^2$ for all $x$ .

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reni_wee

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reni_wee

#44 h Sep 21, 2024, 5:53 PM

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$f(x^2+y)=f(f(x)-y)+4f(x)y$ Asserting, $x=y=0$
$\rightarrow f(0)=f(f(0))$ Let $f(0)=t$ , for some $t\in\mathbb{R}$ . $\rightarrow f(t)=t$

Asserting, $x=0, y=t$
$t=t+4t^2\rightarrow t=0$ $\therefore f(0)=0, \hspace{0.2cm} f(x)=f(-x) \rightarrow \text{$f$ is an even function.}$ Asserting $y=-x^2$
$f(f(x)+x^2)=4f(x)x^2$ Asserting $y=f(x)$
$f(f(x)+x^2)=4f(x)^2$ $\rightarrow f(x)(x^2-f(x)=0$ .
$\therefore f(x)=0 \text{ or } f(x)=x^2$ Now, we have to check whether both of these can happen together.

Let $a,b\in\mathbb{R}$ such that, $f(a)=0,\hspace{0.1cm} f(b)=b^2$

Plugging $x=a,\hspace{0.1cm} y=b$
$f(a^2+b)=f(-b)=f(b)=b^2$ $f(a^2+b) = 0 \rightarrow b = 0$ gives us a contradiction.

Hence $f(a^2+b)=a^4+2a^2b+b^2=b^2\rightarrow a^2(a^2+2b)=0$ .

$a=0$ gives us a contradiction. \therefore \hspace{0.1cm} $a^2=-2b$ .

Plugging $x=a,\hspace{0.1cm} y=-b$
$f(-3b)=f(3b)=b^2\rightarrow b=0 \text { which is a contradiction.}$
\therefore\hspace{0.2cm} $f(x)=x^2 \text{ and } f(x)=0$ are the only solutions.

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bjump

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bjump

#45 h Sep 26, 2024, 5:42 PM

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Setting $y=-x^2$ gives
$f(0)=f(f(x)+x^2)-4x^2f(x)$ Setting $y=f(x)$ gives
$f(x^2+f(x))=f(0)+4f(x)^2$ Now
$4f(x)^2 = f(x^2+f(x))-f(0) 4x^2f(x)$ $0=f(x)(x^2-f(x))$ So $f(x) \in \{0, x^2 \}, \forall x \in \mathbb R$ . Now take an $x \neq 0$ and suppose $f(x)=x^2$ then :
$f(x^2+y) = f(x^2-y) + 4x^2y$ Let $y=x^2-a$ for some $x^2 \neq a \neq 0$ and suppose $f(a) = 0$
$f(2x^2-a) = f(a) + 4x^2(x^2-a)$ $f(2x^2-a) = 4x^4-4x^2a$ The RHS is nonzero so $f(2x^2-a) = (2x^2-a)^2 = 4x^3-4x^2a + a^2 = 4x^3-4x^2a$ , or $a=0$ a contradiction. So $f(x)=0$ or $f(x)=x^2$ for all $x$ .

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eg4334

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eg4334

#46 h Feb 17, 2025, 3:23 AM • 1 Y

Y by alexanderhamilton124

Force set equal so consider $P(x, \frac{f(x)-x^2}{2})$ to obtain that $f(x) \in \{0, x^2 \}$ for all $x$ . To solve the pointwise, consisder FTSOC some nonzero $a, b$ such that $f(a)=0$ and $f(b)=b^2$ . Consider $P(a, -b)$ to get a contradiction. Thus, $\boxed{f(x) \equiv 0, x^2}$ .

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raghu7

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raghu7

#47 h Feb 17, 2025, 7:34 AM

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efang wrote:

$f(f(x)+y) = f(x^2-y) + 4f(x)y$

Plug in $y=x=0$ to get $f(f(0)) = f(0)$

Plug in $x=0, y=-f(0)$ to get $f(0) = f(f(0)) - 4f(0)^2$ . Thus $4f(0)^2 = 0$ or $f(0) = 0$

Plug in $y=-f(x)$ we have $f(x^2+f(x)) - 4f(x)^2 = 0$ or $f(x^2+f(x)) = 4f(x)^2$ .

Plug in $y=x^2$ to get $f(f(x)+x^2) = 4x^2f(x)$

Thus $4x^2f(x) = 4f(x)^2$ . So either $f(x) \equiv 0$ or $4f(x) = 4x^2 \implies f(x) \equiv x^2$

This does not complete the solution as it could be

pco wrote:

$0$ (or) $x^2$ for each value of $x\in\mathbb{R}$ .

In order to clarify this by continuing with efang's solution, assume that $a, b\neq 0$ , such that $f(a)\neq 0$ and $f(a^2-b) = 0$ then according to efang's solution $f(a)=a^2$ as the $f$ function can only take either $0$ (or) $x^2$ . Thus, $f(f(x)+y) = f(x^2-y) + 4f(x)y\Rightarrow f(a^2+b) = 4a^2b\neq 0\Rightarrow f(a^2+b) = (a^2+b)^2 = 4a^2b\Rightarrow (a^2-b)^2=0\Rightarrow b = a^2$ . Thus, we can conclude that if there exist $a\neq 0$ such that $f(a)=a^2$ , then $f(y)=0\iff y=0$ and hence $f(x)=x^2,\ \forall x\in\mathbb{R}\backslash\{a^2\}$ . Ofcourse, we need to check the value of $f(a^2)$ for we assumed that $b\neq 0$ . Clearly, $x = a^2, y=0\Rightarrow f(f(x))=f(x^2)\Rightarrow f(f(a^2)) = f(a^4) = a^8\neq 0\Rightarrow f(a^2)\neq 0\Rightarrow f(a^2)=a^4$ . Thus, either $\boxed{f(x)=0,\ \forall x\in\mathbb{R}}$ (or) $\boxed{f(x)=x^2,\ \forall x\in\mathbb{R}}$ .

This post has been edited 1 time. Last edited by raghu7, Feb 17, 2025, 7:40 AM
Reason: fixed few errors and coding mistakes

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quantam13

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quantam13

#48 h Feb 23, 2025, 9:37 AM

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The only solution is $\boxed{f(x)=x^2 \text{ or }f(x)=0}$ which clearly works. Plugging in $y=\frac{f(x)-x^2}{2}$ to get that $f(x)\in \{0,x^2\}$ . But from here, its easy to get the conclusion by showing that $f(1)=1\implies f(x)=x^2$ and that $f(1)=0\implies f\equiv 0$ .

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blueprimes

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blueprimes

#49 h Feb 25, 2025, 1:22 AM

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We claim the answer is $f(x) \equiv 0, x^2$ , both of which can be verified to work. Now we show these are the only ones.

Asserting $y = \dfrac{x^2 - f(x)}{2}$ simplifies the equation to $f(x)(x^2 - f(x)) = 0$ so $f(x) \in \{0, x^2 \}$ for all $x$ . To deal with the pointwise trap, let $a, b$ be nonnegative reals satisfying $f(a) = 0, f(b) = b^2$ . Plugging in $(x, y) = (a, 0)$ yields $f(f(a)) = f(a^2) = 0$ . Now the assertions $(x, y) = (a, b)$ yields
$f(b) = f(a^2 - b) \implies b^2 = (a^2 - b)^2 \implies a^2 = 2b.$ Similarly $(x, y) = (a^2, b)$ yields $a^4 = 2b$ which forces $a = \pm 1$ . Hence, for all $x \ne -1, 1$ we have $f(x) = x^2$ .

$\textbf{Case 1:}$ $a = 1$
So $f(1) = 0$ , the assertion $(x, y) = (1, 3)$ yields $f(f(1) + 3) = f(-2) + 12 f(1) \implies 3^2 = (-2)^2$ which fails.

$\textbf{Case 2:}$ $a = -1$
So $f(-1) = 0$ , the assertion $(x, y) = (-1, 3)$ gives $f(f(-1) + 3) = f(-2) + 12 f(-1) \implies 3^2 = (-2)^2$ which also fails.

We have a contradiction, so $f(x) \equiv 0, x^2$ as desired.

This post has been edited 1 time. Last edited by blueprimes, Feb 25, 2025, 10:43 PM

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Marcus_Zhang

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Marcus_Zhang

#50 h Mar 11, 2025, 3:44 AM

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Pointwise case.

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TheMathematics

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TheMathematics

#51 h Mar 11, 2025, 8:38 AM

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$P(x,0) \rightarrow f(f(x))=f(x^2)$ and $P(0,-f(0)) \rightarrow f(0)=f(f(0)) - 4f(0)^2 \rightarrow f(0)=0$
So $P(x,x^2-f(x)) \rightarrow f(x^2)=f(f(x)) + 4f(x)(x^2-f(x)) \rightarrow f(x)=0 \lor f(x)=x^2$

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pco

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pco

#52 h Mar 11, 2025, 11:18 AM

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TheMathematics wrote:

$P(x,0) \rightarrow f(f(x))=f(x^2)$ and $P(0,-f(0)) \rightarrow f(0)=f(f(0)) - 4f(0)^2 \rightarrow f(0)=0$
So $P(x,x^2-f(x)) \rightarrow f(x^2)=f(f(x)) + 4f(x)(x^2-f(x)) \rightarrow f(x)=0 \lor f(x)=x^2$

You should read the 44 posts above yours. They all explain why you are wrong and what is the pointwise trap you did not avoid.

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Maximilian113

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Maximilian113

#53 h Mar 15, 2025, 11:36 PM

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Let $P(x, y)$ denote the assertion. Then $P(x, f(x))-P(x, -x^2) \implies x^2f(x)=f(x)^2.$ If $x$ is such that $f(x) \neq 0,$ then $f(x)=x^2.$

Now for the sake of a contradiction assume that there are nonzero $a, b$ such that $f(a)=0, f(b)=b^2.$ Note that $P(0, x) \implies f(x)=f(-x).$ Then $P(a, b) \implies f(a^2+b)=f(-b)=f(b) = b^2,$ so $f(a^2+b) > 0 \implies f(a^2+b)=(a^2+b)^2 \implies a^4+2a^2b=0 \implies a^2=-2b.$ As $f(b)=f(-b),$ WLOG $b > 0,$ contradiction.

Hence the only solutions are $f(x) \equiv x^2$ or $f(x) \equiv 0.$

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lpieleanu

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lpieleanu

#54 h Mar 26, 2025, 11:34 AM

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Solution

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