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KevinYang2.71

392 posts

KevinYang2.71

#1 h Mar 21, 2024, 4:00 AM • 7 Y

Y by peelybonehead, megarnie, Rounak_iitr, Blue_banana4, buddyram, deduck, ItsBesi

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

Proposed by Carl Schildkraut

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eibc

595 posts

eibc

#2 h Mar 21, 2024, 4:01 AM • 3 Y

Y by Aryan_Agarwal, aidan0626, Jack_w

Solution avoiding pointwise trap:

The solutions are $f \equiv 0, f \equiv x^2, f \equiv -x^2$ . We can verify these all work, so now we prove they are the only ones.

Let $P(x, y)$ denote the original assertion. From $P(0, 0)$ we have $f(f(0)) = 0$ . Then, from $P(x, \tfrac{x^2}{2})$ we get $x^2f(x) = f(f(x))$ .

Claim 1: $f$ is even

Proof: Substituting $x^2f(x) = f(f(x))$ into the original equation, we have
$f(x^2 - y) + 2yf(x) = x^2f(x) + f(y).$ Then, substituting $x \mapsto -x$ yields
$f(x^2 - y) + 2yf(-x) = x^2f(-x) + f(y).$ Subtracting the two equations, we have
$x^2f(x) - x^2f(-x) - 2yf(x) + 2yf(-x) = 0 \implies (f(x) - f(-x))(x^2 - 2y) = 0,$ so taking $y$ such that $x^2 \neq 2y$ gives $f(x) = f(-x)$ . $\Box$

Now, taking $P(0, y)$ for any nonzero $y$ yields $2y f(0) = 0$ , so $f(0) = 0$ . Then, from $P(x, 0)$ , we have $f(x^2) = f(f(x))$ .

Claim 2: There exists a constant $k$ such that $f(x) = kx^2$ for all $x$

Proof: Since $f$ is even and $f(0) = k \cdot 0^2$ for any constant $k$ , it suffices to prove the claim for positive $x$ . Notice that $xf(\sqrt{x}) = f(f(\sqrt{x})) = f(x)$ , so from $P(\sqrt{x}, y)$ for positive $y$ we have
$f(x - y) + \frac{2yf(x)}{x} = f(x) + f(y).$ Similarly, $P(\sqrt{y}, x)$ yields
$f(y - x) + \frac{2xf(y)}{y} = f(y) + f(x).$ Since $f$ is even, we have $f(x - y) = f(y - x)$ , so $\tfrac{2yf(x)}{x} = \tfrac{2xf(y)}{y} \implies \tfrac{f(x)}{x^2} = \tfrac{f(y)}{y^2}$ , and from varying $y$ it follows that $f(x) = kx^2$ for some constant $k$ . $\Box$

Now, notice that
$kx^4 = f(x^2) = f(f(x)) = k^3x^4,$ so by taking nonzero $x$ we have $k^3 = k \implies k \in \{-1, 0, 1\}$ , which implies the solution set. $\blacksquare$

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OronSH

1718 posts

OronSH

#3 h Mar 21, 2024, 4:01 AM • 1 Y

Y by Aryan_Agarwal

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

We claim the solutions are $f(x)=0,f(x)=x^2,f(x)=-x^2.$

Denote the assertion as $P(x,y).$
Take $P(x,0)$ to get $f(x^2)=f(f(x))+f(0).$
Take $P(x,x^2)$ to get $f(0)+2x^2f(x)=f(f(x))+f(x)^2=2f(f(x))+f(0).$ Thus $x^2f(x)=f(f(x)).$ In particular, $f(f(0))=0.$
From $P(x,0)$ and $P(-x,0)$ we see $f(f(x))=f(f(-x)).$ Thus $x^2f(x)=x^2f(-x),$ so for $x\ne 0$ we have $f(x)=f(-x)$ and thus $f$ is even.
Take $P(0,y)$ to get $f(-y)+2yf(0)=f(f(0))+f(y).$ This simplifies to $2yf(0)=0,$ so $f(0)=0.$

Now suppose $f(k)=0.$ From $P(k,0)$ we get $f(k^2)=0.$ Taking $P(k,x^2)$ gives $f(x^2)=f(k^2-x^2)=f(x^2-k^2).$ Then $P(x,k^2)$ gives $f(x^2-k^2)+2kf(x)=f(f(x))=f(x^2),$ so $2kf(x)=0,$ so either $f$ is identically zero or $k=0.$

Now suppose $f(a)=f(b)\ne 0.$ From $x^2f(x)=f(f(x))$ we have $a^2f(a)=f(f(a))=f(f(b))=b^2f(b)$ so $a^2=b^2$ and $a=\pm b.$ From $P(x,0)$ we get $f(x)=\pm x^2$ for all $x.$

Taking $P(x,f(y))$ gives $f(x^2-f(y))=f(f(x))+f(f(y))-2f(x)f(y).$ This is symmetric, so $f(x^2-f(y))=f(y^2-f(x))$ so $x^2-f(y)=\pm(y^2-f(x)).$

Now suppose $f(a)=a^2,f(b)=-b^2.$ Taking $x=a,y=b$ in the above, we get $a^2+b^2=\pm(b^2-a^2),$ but clearly this only holds when either $a=0$ or $b=0.$

Thus the only possible solutions are $f(x)=0,f(x)=x^2,f(x)=-x^2$ and we may easily check that these work.

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GrantStar

812 posts

GrantStar

#4 h Mar 21, 2024, 4:02 AM • 1 Y

Y by Jack_w

how many points if i got $x^6f(x)=x^2f(x)^3$ and said this means $f(x)=0,x^2$ and did pointwise trap on these ( i forgot about $-x^2$ )

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MathLuis

1448 posts

MathLuis

#5 h Mar 21, 2024, 4:02 AM • 2 Y

Y by ACBY, KevinYang2.71

Let $P(x,y)$ the assertion of the following F.E., i claim that $f(x)=0,x^2,-x^2$ work.
Indeed it is trivial by replacing, now we prove those are the only ones that work.
$P(0,0)$ gives $f(f(0))=0$ , $P(x,f(y))$ with symetry gives $f(x^2-f(y))=f(y^2-f(x))$ call this $Q(x,y)$ .
Now $P(0,x)$ gives $f(-x)-f(x)=-2xf(0)$ , $P(x,0)$ gives $f(x^2)=f(f(x))+f(0)$ which gives $f (f(x))=f(f(-x))$ .
Also $P(-x,y)$ for $y \ne 0$ gives $2yf(x)=2yf(-x)$ therefore $f$ is even and thus $f (0)=0$ .
Now we have $f(x^2)=f(f(x))$ as well, if $f(a)=f(b)$ y $P(a,x)-P(b,x)$ we get $f(a^2-x)=f (b^2-x)$ .
Therefore if $f$ is periodic (say its period is $T$ ), by $P(x,y)-P(x,y+T)$ we get $2yf(x)=2 (y+T)f(x)$ .
From here we can conclude that $f(x)=0$ or $a^2=b^2$ , suppose the latter then $f(x)=x^2$ or $f(x)=-x^2$ .
Suppose $f(a)=a^2$ and $f(b)=-b^2$ for $a,b \ne 0$ , then by $Q(a,b)$ we get $f(a^2+b^ 2)=f(a^2-b^2)$
If $f(x)=x^2$ for both then $b=0$ , if $f(x)=-x^2$ then $b=0$ so suppose $(a^2+b^2)^2=-(a ^2-b^2)^2$ , then $a=b$ which cannot happen, therefore $f(x)=x^2$ or $f(x)=-x^2$ as desired thus we are done :cool:

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EpicBird08

1730 posts

EpicBird08

#6 h Mar 21, 2024, 4:07 AM • 2 Y

Y by WizardOfChess2011, Tem8

Here is my cursed solution which somehow got a 7/7.

The only solutions are $\boxed{f(x) = 0}, \boxed{f(x) = x^2},$ and $\boxed{f(x) = -x^2}.$ These can all be sought out to satisfy our functional equation.

Now we show that these are the only solutions. Let $P(x,y)$ denote the given equation.

$P\left(x,\frac{x^2}{2}\right)$ gives
$f(f(x)) = x^2 f(x).$ $P(x,0)$ gives
$f(x^2) = f(f(x)) + f(0).$ On the other hand, $P(-x,0)$ gives
$f(x^2) = f(f(-x)) + f(0).$ Combining these two gives
$f(f(-x)) = f(f(x)).$ Using $f(f(x)) = x^2 f(x)$ gives
$x^2 f(x) = x^2 f(-x) \rightarrow \boxed{f(x) = f(-x) \text{ for all } x}.$ Then $P(0,x)$ gives
$f(-x) + 2x f(0) = f(f(0)) + f(x).$ Since $f(f(0)) = f(0) \cdot 0^2 = 0$ and $f(-x) = f(x),$ we get $2x f(0) = 0,$ so $f(0) = 0.$ Plugging into $f(x^2) = f(f(x)) + f(0)$ thus yields
$\boxed{f(f(x)) = f(x^2) = x^2 f(x)}.$ We now find the value of $f(f(f(x)))$ in two different ways. The first is to use the above identity on $f(x)$ to get
$f(f(f(x)) = f(x)^2 f(f(x)) = f(x)^2 \cdot x^2 \cdot f(x) = f(x)^3 \cdot x^2.$ The second way is to find $f(f(x))$ and stick it into $f.$ We get
$f(f(f(x)) = f(f(x^2)) = (x^2)^2 f(x^2) = x^4 \cdot x^2 \cdot f(x) = x^6 \cdot f(x).$ Therefore,
$x^6 \cdot f(x) = x^2 \cdot f(x)^3.$ This implies that $f(x) = 0, x^2, -x^2$ for all individual $x.$

We now deal with the pointwise trap. There are $3$ cases that we consider:

Case 1: $f(a) = a^2$ and $f(b) = 0$ for nonzero $a$ and $b.$ Then $P(b,a)$ yields
$f(b^2 - a) + 2a f(b) = f(f(b)) + f(a) \rightarrow f(b^2 - a) = f(0) + a^2 = a^2,$ while $P(b,-a)$ yields
$f(b^2+a) - 2a f(b) = f(f(b)) + f(-a) \rightarrow f(b^2 + a) = 0 + f(a) = a^2.$ Since $a^2 > 0,$ we require that $f(b^2 - a) = (b^2 - a)^2 = a^2$ and $f(b^2+a) = (b^2 + a)^2 = a^2.$ Thus $(b^2 - a)^2 = (b^2 + a)^2,$ which simplifies down to $4a^2 b = 0.$ This is impossible since $a,b \ne 0.$

Case 2: $f(a) = -a^2$ and $f(b) = 0$ for nonzero $a$ and $b.$ Then $P(b,a)$ yields
$f(b^2 - a) + 2a f(b) = f(f(b)) + f(a) \rightarrow f(b^2 - a) = f(0) - a^2 = -a^2,$ while $P(b,-a)$ yields
$f(b^2+a) - 2a f(b) = f(f(b)) + f(-a) \rightarrow f(b^2 + a) = 0 + f(a) = -a^2.$ Since $-a^2 < 0,$ we require that $f(b^2 - a) = -(b^2 - a)^2 = -a^2$ and $f(b^2+a) = -(b^2 + a)^2 =- a^2.$ Thus $(b^2 - a)^2 = (b^2 + a)^2,$ which simplifies down to $4a^2 b = 0.$ This is impossible since $a,b \ne 0.$

Case 3: $f(a) = a^2$ and $f(b) = -b^2$ for nonzero $a$ and $b.$ Then $P(a,b)$ yields
$f(a^2 - b) + 2b f(a) = f(f(a)) + f(b) \rightarrow f(a^2-b) = f(a^2) - 2ba^2 - b^2 = a^2 f(a) - 2ba^2 - b^2 = a^4 - 2ba^2 - b^2$ while $P(a,-b)$ yields
$f(a^2 + b) - 2b f(a) = f(f(a)) + f(-b) \rightarrow f(a^2+b) = f(a^2) + 2ba^2 - b^2 = a^4 + 2ba^2 - b^2.$ If both of these are equal to $0,$ this clearly implies that $4ba^2 = 0,$ which is impossible. Thus there are $4$ subcases from here:

Subcase 3a: $f(a^2 - b) = (a^2-b)^2 = a^4 - 2ba^2 - b^2.$ This simplifies to $2b^2 = 0,$ impossible.

Subcase 3b: $f(a^2 + b) = (a^2 + b)^2 = a^4 + 2ba^2 - b^2.$ This also yields $2b^2 = 0,$ impossible.

Subcase 3c: $f(a^2 - b) = -(a^2 - b)^2 = a^4 - 2ba^2 - b^2.$ This simplifies to $2a^4 - 4ba^2 = 0,$ so $a^2 = 2b.$ Plugging into $P(a,-b)$ gives
$f(3b) = (2b)^2 + 2b(2b) - b^2 = 7b^2 \ne 0, 9b^2, -9b^2,$ which is a contradiction.

Subcase 3d: $f(a^2 + b) = -(a^2 + b)^2 = a^4 + 2ba^2 - b^2.$ This yields $2a^4 + 4ba^2 = 0,$ so $a^2 = -2b.$ Plugging into $P(a,b)$ gives
$f(-3b) = (-2b)^2 - 2b(-2b) - b^2 = 7b^2 \ne 0, 9b^2, -9b^2,$ which is impossible.

We have exhausted through all cases, so Case $3$ is impossible.

Therefore, the pointwise trap is impossible, and so the only solutions to the given functional equation are those given at the beginning of the solution, as desired.

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pi_is_3.14

1437 posts

pi_is_3.14

#7 h Mar 21, 2024, 4:08 AM

Y by

Here is my solution: no pointwise

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Technodoggo

1928 posts

Technodoggo

#8 h Mar 21, 2024, 4:10 AM

Y by

How many points for getting the right answer and most of the proof??

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krithikrokcs

142 posts

krithikrokcs

#9 h Mar 21, 2024, 4:10 AM

Y by

I was nearly trapped by pointwise and fixed it in last 15 min; answer is still $f(x)=x^2,-x^2,0$

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Orthogonal.

585 posts

Orthogonal.

#10 h Mar 21, 2024, 4:10 AM

Y by

how many points for doing everything like post #5 but only providing like 70 percent detail in the proof of no pointwise trap

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Kevin_Liu

33 posts

Kevin_Liu

#11 h Mar 21, 2024, 4:13 AM • 2 Y

Y by Professor33, elasticwealth

oopsy daisies

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Reason: dumb q

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Mathandski

706 posts

Mathandski

#12 h Mar 21, 2024, 4:18 AM

Y by

Solution Sketch:

P(0, 0) -> f(f(0)) = 0
Also get,
f(x^2) = x^2 f(x) = f(f(x))
So f(0) = 0

$f$ is even from taking $x = 0$ .

If f(n) = 0 for some $n \neq 0$ , we plug in $x = n$ to get,
$f(n^2 - y) = f(y)$
Then, plug in $y = x^2 - n^2$ to eventually get $-2n^2 f(x) = 0 \iff f(x) = 0$ for all $x$ .

Prove injectivity over positives by FTSOC f(m) = f(n) but $m \neq n > 0$ ...

Now look at $f(f(x)) = f(x^2)$ to get $f(x) = \pm x^2$ depending on its sign (use even).

$f(x) \in {x^2, -x^2}$ pretty straight-forward pointwise trap.

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Danielzh

476 posts

Danielzh

#13 h Mar 21, 2024, 4:28 AM

Y by

~~how many points for showing~~ $x^6*f(x)=x^2f(x)^3$ :what?:

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Technodoggo

1928 posts

Technodoggo

#14 h Mar 21, 2024, 4:52 AM

Y by

I showed that $f(f(x))=f(x^2)$ , and I wrote that $f(x)=x^2$ , $0$ , or $-x^2$ . I also made a reference to the pointwise trap and wrote a little bit of garbage about avoiding it; I think it's obvious enough that if you have $f(a)=a^2$ and $f(b)=0$ or $f(a)=a^2$ and $f(b)=-b^2$ or $f(a)=-a^2$ and $f(b)=0$ , then it doesn't satisfy the original equation, but I don't think I made that clear enough since I had literally 1 minute left while writing that. :oops:

I also showed that $f(0)=0$ . I talked about how if $f(x)$ was injective, then $f(x)=x^2$ , but clearly $f$ is not injective since we have $f(x)=f(-x)$ . I couldn't figure out how to rigorously prove that $f$ could only be $x^2$ , $0$ , and $-x^2$ though :noo:

hoping for a 5 or somehow 6 on this one :noo:

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gracemoon124

872 posts

gracemoon124

#15 h Mar 21, 2024, 4:54 AM • 1 Y

Y by Jack_w

i missed -x^2 help

but i proved f(0)=0, f(1)=0 or 1, f(x^2-x)=(x-1)^2f(x), f(x)=f(-x), and f(x^2)=x^2f(x)=f(x^2). how many points would i get?

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KevinChen_Yay

185 posts

KevinChen_Yay

#66 h May 1, 2024, 7:51 PM

Y by

Awesomeness_in_a_bun wrote:

they gave 0 for just including the functions

Yea I wrote the correct answer along with full proof, only error being assuming polynomial and got a 0 rips

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OptimalFEian

10 posts

OptimalFEian

#67 h Jul 3, 2024, 3:45 AM

Y by

It is obvious that the functions $f(x) = x^2 \thinspace \forall x \in \mathbb{R}$ and $f(x)\equiv 0$ satisfy the given equation. Now, we will prove that these are the only solutions. Let $P(x, y)$ denote the given assertion.
$\begin{align*} P(0, 0)&: f(f(0)) = 0 \\ P(f(0), f(0)^2/2)&: f(f(0)^2/2) = f(0) + f(f(0)^2/2) \implies f(0) = 0 \\ P(x, 0)&: f(x^2) = f(f(x)) \\ P(0, x)&: f(-x) = f(x) \\ P(x, x^2/2)&: f(f(x)) = x^2 f(x). \end{align*}$ Then, we can rewrite $P(x, y)$ as
$f(x^2 - y) = (x^2 - 2y) f(x) + f(y).$ Replacing $y$ by $-y^2$ gives
$f(x^2 + y^2) = x^2 f(x) + y^2 f(y) + 2y^2 f(x).$ By symmetry, we get $y^2 f(x) = x^2 f(y)$ . Setting $y \to 1$ , we have $f(x) = cx^2$ for some constant $c$ . Since $f(f(x)) = x^2 f(x)$ , it follows that $c = 0$ or $1$ .

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ihatemath123

3426 posts

ihatemath123

#70 h Aug 16, 2024, 3:09 AM

Y by

The solutions are $f(x) = x^2$ , $f(x) = -x^2$ and $f(x) = 0$ .

Claim: We have $f(0) = 0$ .
Proof: The claim follows if we subtract $P(1, \tfrac{1}{2} )$ from $P(1,1)$ .

Now, taking $P(x,0)$ gives us $f(x^2) = f(f(x))$ , so our FE is equivalent to
$f(x^2 - y) = (x^2 - 2y) f(x) + f(y).$ Let $Q(x,y)$ denote the above assertion.

Claim: If there exists a nonzero root of $f$ , then $f(x) = 0$ for all $x$ .
Proof: Let this nonzero root be $r$ . Subtracting $Q(0,x)$ from $Q(r,x)$ gives us
$f(r^2 - x) = f(-x),$ so $f$ is periodic with period $r^2$ . Now, subtracting $Q(x, r^2)$ from $Q(x,0)$ gives us $r^2 f(x) = 0$ for all $x$ , so $f$ is identically $0$ .

From hereon, assume $0$ is the only root of $f$ . Taking $P(x, \tfrac{x^2}{2} )$ where $x \neq 0$ gives us $x^2 f(x) = f(f(x))$ , implying "quasi-injectivity": that if $f(a) = f(b)$ , then $|a| = |b|$ . Now, taking $P(x,0)$ gives us $f(x^2) = f(f(x))$ , so $f(x) = \pm x^2$ for each $x$ .

Claim: Either $f(x) = x^2$ for all $x$ or $f(x) = -x^2$ for all $x$ .
Proof: Suppose otherwise. Notice that if $f$ is a solution to our rewritten FE, then $-f$ is also a solution. So, assume WLOG that $f(1) = 1$ . Now, let $x$ and $y$ be nonzero reals with such that $x^2 - y = 1$ :

If, FTSOC, $f(x) = x^2$ and $f(y) = -y^2$ , taking $Q(x,y)$ gives us $y = 0$ , which is a contradiction.
If, FTSOC, $f(x) = -x^2$ and $f(y) = y^2$ , taking $Q(x,y)$ gives us
$x^2(x^2 - 2y) = 0 \implies y = x^2 - y = 1 \implies x = \sqrt{2}.$ But now, taking $Q(1, \sqrt{2})$ gives us $\left| (1 - \sqrt{2})^2 \right| = \left| -1 - 2 \sqrt{2} \right|$ , contradiction.
If, FTSOC, $f(x) = -x^2$ and $f(y) = -y^2$ , taking $Q(x,y)$ gives us $2(x^2 - y)^2 = 0$ , contradiction.

It follows that $f(x) = x^2$ or $f(x) = -x^2$ .

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CyclicISLscelesTrapezoid

371 posts

CyclicISLscelesTrapezoid

#71 h Sep 1, 2024, 9:15 PM • 4 Y

Y by GrantStar, OronSH, megarnie, centslordm

The solutions are $f(x)=0$ , $f(x) \equiv x$ , and $f(x) \equiv -x^2$ , which work. Let $P(x,y)$ denote the assertion in the problem statement.

$P(0,0)$ gives $f(f(0))=0$ .
$P(f(0),\tfrac{f(0)^2}{2})$ gives $f(0)=0$ .
$P(0,x)$ gives $f(x)=f(-x)$ .
$P(x,0)$ gives $f(f(x))=f(x^2)$ .

$P(x,y^2)$ gives
$f(x^2-y^2)+2y^2f(x)=f(f(x))+f(y^2)=f(x^2)+f(y^2).$ Swapping $x$ and $y$ gives
$f(x^2-y^2)+2y^2f(x)=f(y^2-x^2)+2x^2f(y)=f(x^2-y^2)+2x^2f(y),$ so $y^2f(x)=x^2f(y)$ . Plugging in $y=1$ gives $f(x)=f(1)x^2$ .

Since $f(f(x))=f(x^2)$ , we have
$f(1)(f(1)x^2)^2=f(1)x^4 \implies f(1)^3=f(1),$ so $f(1) \in \{0,1,-1\}$ , as desired. $\blacksquare$

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fearsum_fyz

48 posts

fearsum_fyz

#72 h Oct 24, 2024, 9:01 AM

Y by

We claim the only answers are $\boxed{f(x) = 0}$ , $\boxed{f(x) = x^2}$ , and $\boxed{f(x) = - x^2}$ . It is easy to verify that these work. It remains to show that they the only solutions.

Claim 1: $x^2 f(x) = f(f(x))$
Proof.
$\underline{P}(x, 0) \implies f(x^2) = f(f(x)) + f(0)$
$\underline{P}(x, x^2) \implies f(0) + 2x^2f(x) = f(f(x)) + f(x^2)$
Adding, we get $2x^2 f(x) = 2f(f(x)) \implies \boxed{x^2 f(x) = f(f(x))}$

Claim 2: $f(0) = 0$
Proof.
By Claim 1, we have $f(f(1)) = f(1)$ . Therefore:
$\underline{P}(1, 1) \implies f(0) + \cancel{2f(1)} = \cancel{2f(1)}$
$\implies \boxed{f(0) = 0}$ .

Claim 3: $f$ is even.
Proof.
Using Claim 2:
$\underline{P}(0, y) \implies \boxed{f(-y) = f(y)}$

Claim 4: If $f \not\equiv 0$ , then $f(x) = f(y) \implies x = \pm y$ .
Proof.
By Claim 1, $f(x) = f(y) \implies f(f(x)) = f(f(y)) \implies \frac{f(f(x))}{f(x)} = \frac{f(f(y))}{f(y)} \implies x^2 = y^2 \implies x = \pm y$ .

It is easy to see that Claim 4 then finishes the problem: If $f \not\equiv 0$ , then we have
$f(x^2) = f(f(x)) \implies \boxed{f(x) = \pm x^2}$ .

It remains to take care of the pointwise trap, which is not hard to do. However, for the sake of compactness, I have excluded the details from my post.

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eg4334

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eg4334

#73 h Oct 29, 2024, 4:19 AM • 1 Y

Y by Marcus_Zhang

I claim the only solutions are $\boxed{f(x) \equiv 0, x^2, -x^2}$ which can easily be verified to work. Let $P(x, y)$ denote the given assertion.

$P(x, \frac{x^2}{2})$ gives $f(f(x)) = x^2f(x)$ .

$P(-x, y)$ and $P(x, y)$ and equating gives $x^2f(x)-2yf(x)=x^2f(-x)-2yf(-x)$ . $y=0$ so $x^2f(x)=x^2f(-x)$ or $f(x)=f(-x)$ for nonzero $x$ and thus for all $x$ . Therefore $f$ is even. Call this statement $1$ .

Let $x=0$ immediately gives us $2yf(0)=0$ so $f(0)=0$ , coupled with the fact that $f(f(0))=0$ . Now let $y=0$ , so $f(x^2)=f(f(x))$ , and we have the chain: $f(x^2)=f(f(x))=x^2f(x)$ which is of tremendous use. Let $x=f(x)$ in the latter inequality to get $f(f(f(x)))=f(x)^2f(f(x))$ $f(x)^2 f(x^2) = x^4 f(x^2)$ so $f(x) = 0, x^2, -x^2$ as mentioned before for an individual $x$ .

We now tackle the pointwise trap. I provide a sketch below.

If $f(a)=a^2$ and $f(b)=0$ for $a, b \neq 0$ , then consider $P(b, a)$ and $P(b, -a)$ . This readily shows $f(b^2-a) = f(b^2+a)=a^2$ . This contradicts $a, b \neq 0$ .

If $f(a)=-a^2$ and $f(b)=0$ for $a, b \neq 0$ then consider $P(b, a)$ and $P(b, -a)$ yet again. This shows that $(b^2-a)^2 = (b^2+a)^2$ again which contradicts $a, b \neq 0$ .

If $f(a)=a^2$ and $f(b)=-b^2$ then $P(\sqrt{a}, b)$ and $P(\sqrt{b}, a)$ give us another contradiction.

This post has been edited 1 time. Last edited by eg4334, Mar 1, 2025, 1:09 AM

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cappucher

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cappucher

#74 h Dec 23, 2024, 8:16 AM

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The only solutions that satisfy the given are $f(x) \equiv x^2, -x^2, 0$ .

Claim 1: $f(0) = 0$

Proof: Let $P(x, y)$ denote the given equation. From $P(x, 0)$ , we obtain

$f(x^2) = f(f(x)) + f(0)$
Let $c = f(0)$ . From $P(0, 0)$ , we obtain

$c = f(c) + c \implies f(c) = 0 \implies f(f(c)) = c$
From $P(c, 0)$ , we obtain

$f(c^2) = f(f(c)) + c \implies f(c^2) = 2c$
However, from $P(c, c^2)$ , we know that

$f(0) + 2c^2f(c) = f(f(c)) + f(c^2)$ $c = c + f(c^2) \implies f(c^2) = 0$ $f(c^2) = 2c = 0 \implies c = 0$
as desired.

Claim 2: $f(f(x)) = f(x^2) = x^2f(x)$ .

Proof: From $P(x, 0)$ , we have

$f(x^2) = f(f(x)) + f(0) \implies f(x^2) = f(f(x))$
From $P(x, x^2)$ , we have

$f(x^2 - x^2) + 2x^2f(x) = f(f(x)) + f(x^2)$ $2x^2f(x) = 2f(x^2) \implies x^2f(x) = f(x^2)$
as desired.

Claim 3: $f(x)$ is even.

Proof: From $P(0, y)$ , we have

$P(0 - y) + 2yf(0) = f(f(0)) + f(y)$ $f(-y) = f(y)$
as desired.

Using claims $2$ and $3$ , we will prove that $f(x) \equiv x^2, -x^2, 0$ . After we finish doing so, we will show these are the only solutions by avoiding the pointwise trap.

Note that

$f(f(f(x))) = f(f(x^2)) = x^4f(x^2) = x^6f(x)$
and that

$f(f(f(x))) = (f(x))^2f(f(x)) = x^2(f(x))^3$
We can then equate the two to obtain

$x^6f(x) = x^2\left(f(x)\right)^3$ $x^2f(x)\left(f(x)^2 - x^4\right) = 0$ $x^2f(x)\left(f(x) + x^2\right)\left(f(x) - x^2\right) = 0$
The only way for this equation to be satisfied for all $x$ is if $f(x) \equiv -x^2, x^2, 0$ .

Avoiding the pointwise trap

There are three cases to tackle: $f(x) \equiv 0, x^2$ , $f(x) \equiv 0, -x^2$ , and $f(x) \equiv -x^2, x^2$ . Because these take a while to type out, I will edit this solution later to include them. The basic idea is to have non-zero variables $a$ and $b$ such that $f(a) = 0$ and $f(b) = b^2$ (for case 1; same logic applies for cases 2 and 3) and show that this leads to a contradiction with the given. Claim 3 ( $f(x)$ is even) will prove useful for this.

This post has been edited 2 times. Last edited by cappucher, Dec 23, 2024, 8:19 AM

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Yiyj1

1175 posts

Yiyj1

#75 h Dec 23, 2024, 8:25 AM • 1 Y

Y by megarnie

OronSH wrote:

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

We claim the solutions are $f(x)=0,f(x)=x^2,f(x)=-x^2.$

Denote the assertion as $P(x,y).$
Take $P(x,0)$ to get $f(x^2)=f(f(x))+f(0).$
Take $P(x,x^2)$ to get $f(0)+2x^2f(x)=f(f(x))+f(x)^2=2f(f(x))+f(0).$ Thus $x^2f(x)=f(f(x)).$ In particular, $f(f(0))=0.$
From $P(x,0)$ and $P(-x,0)$ we see $f(f(x))=f(f(-x)).$ Thus $x^2f(x)=x^2f(-x),$ so for $x\ne 0$ we have $f(x)=f(-x)$ and thus $f$ is even.
Take $P(0,y)$ to get $f(-y)+2yf(0)=f(f(0))+f(y).$ This simplifies to $2yf(0)=0,$ so $f(0)=0.$

Now suppose $f(k)=0.$ From $P(k,0)$ we get $f(k^2)=0.$ Taking $P(k,x^2)$ gives $f(x^2)=f(k^2-x^2)=f(x^2-k^2).$ Then $P(x,k^2)$ gives $f(x^2-k^2)+2kf(x)=f(f(x))=f(x^2),$ so $2kf(x)=0,$ so either $f$ is identically zero or $k=0.$

Now suppose $f(a)=f(b)\ne 0.$ From $x^2f(x)=f(f(x))$ we have $a^2f(a)=f(f(a))=f(f(b))=b^2f(b)$ so $a^2=b^2$ and $a=\pm b.$ From $P(x,0)$ we get $f(x)=\pm x^2$ for all $x.$

Taking $P(x,f(y))$ gives $f(x^2-f(y))=f(f(x))+f(f(y))-2f(x)f(y).$ This is symmetric, so $f(x^2-f(y))=f(y^2-f(x))$ so $x^2-f(y)=\pm(y^2-f(x)).$

Now suppose $f(a)=a^2,f(b)=-b^2.$ Taking $x=a,y=b$ in the above, we get $a^2+b^2=\pm(b^2-a^2),$ but clearly this only holds when either $a=0$ or $b=0.$

Thus the only possible solutions are $f(x)=0,f(x)=x^2,f(x)=-x^2$ and we may easily check that these work.

Oron orz

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D4N13LCarpenter

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D4N13LCarpenter

#76 h Jan 8, 2025, 5:43 PM • 1 Y

Y by Vahe_Arsenyan

Let $P(x, y)$ denote the given assertion. We claim that the only solutions are $f\equiv0$ , $f(x)=x^2$ and, $f(x)=-x^2$ which clearly work.

Claim 1.4 We have $f(x^2)=f(f(x))$
Proof $P(x,0)$ yields $f(x^2)=f(f(x))+f(0)$ , so it suffices to prove that $f(0)=0$ . Observe that by $P(1,1)$ , we get $f(0)+2f(1)=f(f(1))+f(1)$ but after cancelling out terms and substituting $f(f(1))$ for $f(1)-f(0)$ , we get $f(0)+f(1)=f(1)-f(0)$ which directly implies the result.
Notice how by using $f(0)=0$ we can now easily prove that $f$ is even, as $P(0,y)$ gives $f(-y)=f(y)$ . The key claim is as follows.
Claim 1.5 $f(x)=f(y)$ implies $\vert x\vert=\vert y\vert$ or $f(x)=f(y)=0$
Proof $P(x, y^2)$ states $f(x^2-y^2)+2y^2f(x)=f(f(x))+f(y^2)$ and $P(y, x^2)$ states $f(y^2-x^2)+2x^2f(y)=f(f(x))+f(x^2)$ but using both that $f(x^2)=f(f(x))$ and that $f$ is even, this is equivalent to $y^2f(x)=x^2f(y)=x^2f(x)$ so either $f(x)=0$ or $x^2=y^2\implies \vert x\vert=\vert y\vert y$
Claim 1.6 If $f(c)=0$ for some positive real $c$ , than $f\equiv0$
Proof $P(c,x)$ gives $f(c^2-y)=f(f(c))+f(y)$ but note that $f(f(c))=f(0)=0$ so we in fact have $f(c^2-y)=f(y)$ . However, by Claim 1.5 either $f(c^2-y)=f(y)=0$ or $\vert c^2-y\vert=\vert y\vert$ , the latter being only true when $y=\frac{c^2}{2}$ . Thus, we have $f(x)=0$ for all $x\neq \frac{c^2}{2}$ . But one can now easily find $\frac{c^2}{2}$ taking any $x, y$ satisfying $x^2-y=\frac{c^2}{2}$ , as then $P(x,y)$ gives $f(\frac{c^2}{2})=0$ .
From now on, suppose that $f(c)\neq0$ for any $c$ greater than $0$ . Recall that $f(x^2)=f(f(x))$ so we can apply Claim 1.5 to get $\vert f(x)^2\vert=\vert x^4\vert$ , which yields $f(x)=\pm x^2$ .

We now aim to prove that $f$ can't change of sign, thus leaving us with only the initially stated solutions. Assume for the sake of contradiction that $f$ changes in sign. By Pigeonhole principle, one of these must repeat an infinite number of times, WLOG say it is $f(x)=x^2$ , as the other is completely symmetric. Now choose a sufficiently large $a$ such that $f(a)=a^2$ and a small $b\neq 0$ in comparison which satisfies $f(b)=-b^2$ . Now notice that $P(a,b)$ gives $f(a^2-b)+2a^2b=a^4-b^2$ where we used $f(f(x))=x^2f(x)$ , which is a consequence of $P(x^2, y^2)$ . We can rewrite this as $f(a^2-b)=a^4-b^2-2a^2b$ but we also know $f(a^2-b)=\pm (a^2-b)^2$ . So we can now distinguish two cases.

Case 1. $f(a^2-b)=(a^2-b)^2$
Combining this with $f(a^2-b)=a^4-b^2-2a^2b$ we get $a^4-b^2-2a^2b=a^4-2a^2b+b^2$ so $-b^2=b^2\implies b=0$ . But we supposed $b\neq 0$ , so this case leads to a contradiction.

Case 2. $f(a^2-b)=-(a^2-b)^2$
Notice how $a^4-b^2-2a^2b$ is positive for sufficiently large $a$ , however $-(a^2-b)^2$ is always negative, so we can immediately discard this case.

This post has been edited 2 times. Last edited by D4N13LCarpenter, Jan 10, 2025, 7:02 PM
Reason: I accidentally wrote $f(b)=b^2$ but it's clearly $f(b)=-b^2$

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anandswaroop191

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anandswaroop191

#77 h Jan 9, 2025, 6:18 PM

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Plug in $y = 0$ to get $f(f(x)) = f(x^2) - f(0)$ , substitute in the original equation to get $f(x^2 - y) + 2yf(x) = f(x^2) + f(y) - f(0)$ *.

Plug in $x = y = 1$ to get $f(0) + 2f(1) = f(1) + f(1) - f(0) \implies f(0) = 0$ . Substitute into * to get $f(x^2 - y) + 2yf(x) = f(x^2) + f(y)$ **.

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

Substitute $y \mapsto -y$ to get $f(x^2 + y) - 2yf(x) = f(x^2) + f(-y) = f(x^2) + f(y)$ . Subtract ** from this to get $f(x^2 + y) - f(x^2 - y) = 4yf(x)$ . Substitute $y \mapsto y^2$ to get $f(x^2 + y^2) - f(x^2 - y^2) = 4y^2f(x)$ .

Note that $f(x^2 - y^2) = f(y^2 - x^2)$ (because $f$ is even), so $4y^2f(x) = f(x^2 + y^2) - f(x^2 - y^2) = f(y^2 + x^2) - f(y^2 - x^2) = 4x^2f(y)$ .

Simplifying, we get $y^2f(x) = x^2f(y)$ . Plug in $y = 1$ to get $f(x) = x^2f(1)$ . Therefore, $f(x) = cx^2$ for some constant $c$ .

Going back to the original equation, we have $f(x^2 - y) + 2yf(x) = cx^4 - 2cx^2y + cy^2 + 2x^2y = cx^4 + cy^2$ , and $f(f(x)) + f(y) = c^3x^4 + cy^2$ . Equating these, we have $(c^3 - c)x^4 = 0$ for all real $x$ ; solving $c^3 - c = 0$ , we get $c \in \{-1, 0, 1\}$ .

The desired functions are $f(x) = -x^2$ , $f(x) = 0$ , and $f(x) = x^2$ . $\square$

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mathwiz_1207

91 posts

mathwiz_1207

#78 h Jan 30, 2025, 6:57 AM

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Avoids pointwise I think. We claim the only solutions are $f \equiv \boxed{0, x^2, -x^2}$ . It can be easily verified that these work. We will now prove that these are the only solutions.

Claim: $f(f(x)) = x^2f(x) = f(x^2)$
Proof. $x = x, y = \frac{x^2}{2}$ yields
$f(\frac{x^2}{2})+ x^2f(x) = f(f(x)) + f(\frac{x^2}{2}) \implies f(f(x)) = x^2f(x)$ Now, $x = x, y = x$ gives
$f(x^2 - x) + 2xf(x) = x^2f(x) + f(x)$ $f(x^2 - x) = (x - 1)^2f(x)$ Plugging in $x = 1$ gives $f(0) = 0$ . Now, $x = x, y = 0$ gives us
$f(x^2) = f(f(x)) = x^2f(x)$ so the claim is proven.

We now prove the second key claim, which almost finishes the problem:

Claim 2: $f$ is even.
Proof. By claim $1$ , we have that
$f(x^2) = x^2f(x)$ and that
$f((-x)^2) = x^2f(-x)$ This implies that $x^2f(x) = x^2f( -x)$ . If $x \neq 0$ , we get $f(x) = f(-x)$ . Otherwise, if $x = 0$ , it is trivial that $f(0) = f(-0)$ . Thus, $f$ is even.

Now for the finish. Plugging in $x = x, y = t^2$ yields
$f(x^2 - t^2) + 2t^2f(x) = f(f(x)) + f(t^2) = f(f(x)) + f(f(t))$ Swapping $x$ and $t$ gives
$f(t^2 - x^2) + 2x^2f(t) = f(f(t)) + f(x^2) = f(f(x)) + f(f(t))$ However, since $f$ is even, the first term of both equations are equal, so we may subtract to get
$2x^2f(t) = 2t^2f(x)$ Assume $x, t$ are nonzero, then this means that $\frac{f(x)}{x^2}$ is equal to a constant for all nonzero values of $x$ . This means $f(x) = cx^2$ , and upon substituting, we find $c = 0, 1, -1$ . Clearly, $x = 0$ also satisfies this, so we find $f \equiv 0, x^2, -x^2$ and we are done.

This post has been edited 1 time. Last edited by mathwiz_1207, Jan 30, 2025, 6:59 AM
Reason: skill issue

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bjump

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bjump

#79 h Feb 23, 2025, 10:42 PM • 2 Y

Y by KenWuMath, megarnie

Setting $x=y=0$ gives $f(f(0))=0$ . Setting $x=1$ , and $y=0$ gives $f(1)=f(f(1))+f(0)$ . Setting $x=y=1$ gives $f(0)+f(1)=f(f(1))$ . Subtracting the previous equations we have made gives $f(0)=0$ . Therefore setting $y=0$ gives $f(x^2)=f(f(x))$ . Setting $x=0$ now gives $f(y)=f(-y)$ . Now I claim that $f(x)=g(x)$ is a valid solution to the functional equation iff $f(x)=-g(x)$ is a valid solution. This follows because : $-f(x^2-y)-2yf(x)=-f(-f(x)) -f(y) = -f(f(x))-f(y) \iff f(x^2-y)+2yf(x) = f(f(x)) + f(y).$ Therefore we may assume that $f(1) \ge 0$ . Notice that setting $x=1$ and $y=x^2$ implies that $f(x^2-1)+2x^2f(1)=f(1)$ , and setting $x=x$ , and $y=1$ gives $f(x^2-1)+2f(x)=f(x^2)+f(1)$ subtracting equations gives $f(x)=f(1)x^2$ . We also have from previosly derived properties $f(1)=f(f(1))= f(1)^3$ . Therefore $f(x)=-x^2, 0 , x^2$ .

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Ilikeminecraft

283 posts

Ilikeminecraft

#80 h Mar 14, 2025, 6:42 AM

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I claim that $f(x) = 0, x^2, -x^2$

Let $P(x,y)$ be our assertion. We start by plugging in a lot of values.

$P(x, 0)$ implies that $f(x^2) = f(f(x)) + f(0),$ and $P(x, x^2)$ implies that $f(0) + 2x^2f(x) = f(f(x)) + f(x^2) = f(f(x)) + f(f(x)) + f(0),$ and thus $x^2f(x) = f(f(x)), (1)$ . $P(0, 0)$ implies that $f(f(0)) = 0.$

\begin{lemma}
$f(0) = 0.$
\end{lemma}
\begin{proof}
Let $f(0) = k.$ Thus, we also have that $f(k) = 0.$

Now, we take $P(0, k),$ which tells us that $f(-k) = -2k^2.$ $P(-k, 0)$ gives us that $f(x^2) = f(f(-k)) + k = k^2f(-k) + k = -2k^4 + k$ , but $P(k, 0)$ gives us that $f(x^2) = f(f(k)) + f(0) = 2k.$ Thus, $(2k^3 + 1)k = 0.$

If $k \neq 0,$ we have that $k = -2^{-\frac13}.$ Thus, $f(-k) = -2^{\frac13} = \frac1k.$ We also know that $k^2f(-k) = f(f(-k)) = f(\frac1k),$ and hence $f(\frac1k) = k.$ Thus, $f(f(\frac1k)) = 0,$ but then we also have that $f(f(\frac1k)) = \frac1{k^2}f(\frac1k) = \frac 1k$ by $(1),$ and hence $\frac1k = 0,$ which is a contradiction. Thus, $f(0) = 0.$
\end{proof}

$P(0, x)$ tells us that $f$ is even. Now, $P(x, y^2)$ tells us that $f(x^2 - y^2) + 2y^2f(x) = f(f(x)) + f(y^2) = x^2f(x) + y^2f(y).$ However, by swapping the variables, we get that $f(y^2 - x^2) + 2x^2f(y) = y^2f(y) + x^2f(x).$ By subtracting these two, we get that $2x^2f(y) + x^2f(x) + y^2f(y) = 2y^2f(x) + y^2f(y) + x^2f(x).$ Thus, $\frac{f(x)}{x^2} = \frac{f(y)}{y^2}.$

Let this constant be $c.$ By plugging it in our equation, we see that $c = 1, 0, -1.$ Thus, $f(x) = 0, -x^2, x^2$

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joshualiu315

2513 posts

joshualiu315

#81 h Yesterday at 4:55 AM

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i believe this dodges pointwise trap

The answer is $f(x) = \boxed{0, \pm x^2}$ .

Let the given assertion be $P(x,y)$ . Note that $P(x,0)$ gives

$f(x^2) = f(f(x))+f(0), \quad \bigstar$
and $P(x,\tfrac{x^2}{2})$ gives $x^2f(x) = f(f(x))$ . We plug this into the original equation to get a new assertion:

$Q(x,y): \ f(x^2-y) + 2yf(x) = x^2f(x)+f(y).$
Plugging in $Q(-x,y)$ gives

$f(x^2-y) + 2yf(-x) = x^2f(-x) + f(y).$
Comparing this to $Q(x,y)$ gives

$(x^2-2y)(f(x)-f(-x)) = 0,$
so for $y \neq \tfrac{x^2}{2}$ , we have $f(x) = f(-x)$ for all $x$ ; in other words, $f$ is even. Then, $Q(0,y)$ gives

$2yf(x) = 0,$
which means $f(0) = 0$ . Plugging this into $\bigstar$ yields $f(x^2) = f(f(x)) = x^2f(x)$ . Comparing $Q(x,y^2)$ and $Q(y,x^2)$ gives

$f(x^2-y^2) + 2y^2f(x) = x^2f(x) + f(y^2) = f(x^2)+f(y^2),$ $f(y^2-x^2) + 2x^2f(y) = y^2f(y)+f(x^2) = f(y^2)+f(x^2).$
Since $f(x^2-y^2) = f(y^2-x^2)$ , we have $2y^2f(x) = 2x^2f(y)$ , or that $\tfrac{f(x)}{x^2} = \tfrac{f(y)}{y^2}$ . Setting this equal to a constant yields $f(x) = cx^2$ for some value of $c$ and all values of $x$ . Then,

$cx^4 = f(x^2) = f(f(x)) = c^3x^4 \implies c \in \{-1,0,1\}.$
This gives us our desired solution set.

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Marcus_Zhang

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Marcus_Zhang

#82 h 5 hours ago

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