average FE

The post below has been deleted. Click to close.

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

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

This post has been edited 1 time. Last edited by KevinYang2.71, Mar 21, 2024, 3:13 PM

Z K Y

The post below has been deleted. Click to close.

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

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. $\blacksquare$

This post has been edited 3 times. Last edited by eibc, Mar 21, 2024, 3:21 PM

Z K Y

The post below has been deleted. Click to close.

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

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.

Z K Y

The post below has been deleted. Click to close.

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

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 $-x^2$ )

Z K Y

The post below has been deleted. Click to close.

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

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:

.

Z K Y

The post below has been deleted. Click to close.

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

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.

This post has been edited 2 times. Last edited by EpicBird08, Oct 10, 2024, 4:46 PM

Z K Y

The post below has been deleted. Click to close.

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

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

Attachments:

Z K Y

The post below has been deleted. Click to close.

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

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??

Z K Y

The post below has been deleted. Click to close.

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

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 $f(x)=x^2,-x^2,0$

Z K Y

The post below has been deleted. Click to close.

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

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

This post has been edited 1 time. Last edited by Orthogonal., Apr 29, 2024, 1:20 AM

Z K Y

The post below has been deleted. Click to close.

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

Kevin_Liu

33 posts

Kevin_Liu

#11 h Mar 21, 2024, 4:13 AM • 2 Y

Y by Professor33, elasticwealth

oopsy daisies

This post has been edited 1 time. Last edited by Kevin_Liu, Sep 30, 2024, 8:50 PM
Reason: dumb q

Z K Y

The post below has been deleted. Click to close.

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

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.

Z K Y

The post below has been deleted. Click to close.

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

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?:

This post has been edited 1 time. Last edited by Danielzh, Mar 21, 2024, 4:29 AM

Z K Y

The post below has been deleted. Click to close.

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

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:

Z K Y

The post below has been deleted. Click to close.

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

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?

Z K Y

The post below has been deleted. Click to close.

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

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

Z K Y

The post below has been deleted. Click to close.

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

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 $1$ .

Z K Y

The post below has been deleted. Click to close.

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

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 $f(x) = -x^2$ .

This post has been edited 1 time. Last edited by ihatemath123, Aug 23, 2024, 3:30 AM

Z K Y

The post below has been deleted. Click to close.

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

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. $\blacksquare$

This post has been edited 1 time. Last edited by CyclicISLscelesTrapezoid, Sep 1, 2024, 9:15 PM

Z K Y

The post below has been deleted. Click to close.

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

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.

Z K Y

The post below has been deleted. Click to close.

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

eg4334

598 posts

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

Z K Y

The post below has been deleted. Click to close.

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

cappucher

88 posts

cappucher

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

Y by

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

Z K Y

The post below has been deleted. Click to close.

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

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

Z K Y

The post below has been deleted. Click to close.

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

D4N13LCarpenter

13 posts

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 $f(b)=-b^2$

Z K Y

The post below has been deleted. Click to close.

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

anandswaroop191

14 posts

anandswaroop191

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

Y by

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 $f(x) = x^2$ . $\square$

Z K Y

The post below has been deleted. Click to close.

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

mathwiz_1207

91 posts

mathwiz_1207

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

Y by

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

Z K Y

The post below has been deleted. Click to close.

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

bjump

962 posts

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 $f(x)=-x^2, 0 , x^2$ .

Z K Y

The post below has been deleted. Click to close.

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

Ilikeminecraft

283 posts

Ilikeminecraft

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

Y by

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, $f(x) = 0, -x^2, x^2$

Z K Y

The post below has been deleted. Click to close.

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

joshualiu315

2513 posts

joshualiu315

#81 h Saturday at 4:55 AM

Y by

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.

Z K Y

The post below has been deleted. Click to close.

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

Marcus_Zhang

936 posts

Marcus_Zhang

#82 h Yesterday at 6:06 PM

Y by

No pointwise