I got $f(x)=x, f(x)=0$ but I'm not confident with my solution

Hypernova wrote: I got $f(x)=x, f(x)=0$ but I'm not confident with my solution I've got the same answer too. This was definitely harder than its position..

I used periodity and found cases such that the period is bigger, equal, smaller than 1

Can you post all of the problems?

Very brief solution (Can be wrong) Find $f(0)=0, f(y^2+f(x)-x)=f(y^2)$ Let $f(x)-x=t\neq 0$ and $|t|$ minimum, $t$ is the period If $f(x)\neq0$ then exists some $r$ such that $|f(r)-r|>|t|$ By periodity exists some integer $n$ such that $0<|f(r+nt)-r-nt|<|t|$, contradiction. So no such $t$, so $f(x)=x$ Edit : Trolled

Hypernova wrote : "let $f(x)-x=t \not=0$ and $|t|$ minimum" We cannot do that, since we don't have any sufficient condition on $f$ to do this...

jlv0903 wrote: Hypernova wrote : "let $f(x)-x=t \not=0$ and $|t|$ minimum" We cannot do that, since we don't have any sufficient condition on $f$ to do this... If there is no such $t$ then isn't $f(x)=x$ for all x?

Hypernova wrote: If there is no such $t$ then isn't $f(x)=x$ for all x? I'm not talking about that part. We cannot guarantee the 'minimum' of nonzero $|f(x)-x|$. There is the 'infimum' of such $|t|$, not 'minimum'. And it's possible that the infimum is zero.

Let $P(x,y)$ be the given assertion $f(f(x)-x+y^2)=yf(y)$. If $f$ is injective, then $P(x, 1)$ gives $f(f(x)-x+1)=f(1)$, so $\boxed{f(x) \equiv x}$. Now assume that $f$ is not injective. Our main claim is the following: Claim. If there is a real number $T \neq 0$ such that $f(x+T)=f(x)$ for all sufficiently large $x$, then $f(x+T)=f(x)$ holds for every $x \in \mathbb{R}$. Proof. Suppose that $f(x+T)=f(x)$ for every $x>N$, and WLOG assume that $T>0$. Fix some $z \in \mathbb{R}$ and $s>N$. Note that there is some integer $n$ such that $z+s+nT-f(s) >\min \{N,0 \}$. Now, the assertions $P \left(s+nT, \sqrt{z+s+nT-f(s)} \right)$ and $P \left( s+(n+1)T, \sqrt{z+s+nT-f(s)} \right)$ gives $f(z-T)=\sqrt{z+s+nT-f(s)} f\left(\sqrt{z+s+nT-f(s)}\right)=f(z)$. This proves the claim. $\square$ Let $T$ be the set of real numbers $t$ such that $f(x+t)=f(x)$ for every $x\in \mathbb{R}$. Comparing $P(a,y)$ and $P(b,y)$, we deduce that $f(a)-a-f(b)+b \in T$ for every $a,b \in \mathbb{R}$. Since $f$ is not injective, there exist some $a <b$ such that $f(a)=f(b)$. Let $t=b-a$, then $t=f(a)-a-f(b)+b$, so $t \in T$. Now comparing $P(t, y)$ and $P(t,y+t)$, we get $$(y+t)f(y+t)=f(f(t)-t+(y+t)^2), yf(y)=f(f(t)-t+y^2).$$Plugging $y=\frac{n-t}{2}$ for some integer $n$, then $2y+t$ is a integer, so $f(f(t)-t+(y+t)^2)=f(f(t)-t+y^2)$ holds since $f$ is $t$-periodic. Hence $tf(y)=0$, or $f\left(\frac{n-t}{2} \right) =0$ for every integer $n$. Now, taking $a=\frac{-t}{2}$, and $b=\frac{1-t}{2}$, we get $f(a)-a-f(b)+b=\frac{1}{2} \in T$. Hence $f\left(x+\frac{1}{2} \right)=f(x)$ for every $x \in \mathbb{R}$. By taking $a=f(t)-t+\left(x+\frac{1}{2}\right)^2$ and $b=f(t)-t+x^2$, then $$f(a)-a-f(b)+b=-x-\frac{1}{4}+\left(x+\frac{1}{2}\right) f\left(x+\frac{1}{2}\right)-xf(x)=\frac{1}{2}f(x)-x-\frac{1}{4} \in T.$$Hence $f(x)-2x-\frac{1}{2}$ is also an element of $T$. Since $f(x)-x-f(0) \in T$, so $(f(x)-x-f(0))-\left(f(x)-2x-\frac{1}{2} \right)=x-f(0)+\frac{1}{2}$ is an element of $T$ for every $x$. This proves that $f$ is constant, and we're done!

Cool solution! Too bad that I was on the wrong track during the contest and submitted the wrong solution

Hopefully this works (it doesn't lol) Hypernova wrote: Find all functions $f$ such that $f:\mathbb{R}\rightarrow \mathbb{R}$ and $f(f(x)-x+y^2)=yf(y)$ Solution. We have $f(f(y^2)) = yf(y)$, so $f(f(1)) = f(1)$. Therefore plugging $x = f(1)$ gives $f(y^2) = yf(y)$. This also means $f(0) = 0$ and replacing $y$ by $-y$ gives $f$ is odd. Now, $\boxed{f(x) = x \text{ for all }x}$ is a valid solution. Therefore we'll now assume that there is some $a$ such that $f(a)\neq a$, so \[f(f(a)-a+y^2) = yf(y) = f(y^2),\]so $f(x+c) = f(x)$ for $x\geq 0$ and some constant $c:=f(a) - a$. Plugging $-a$ gives $f(x-c) = f(x)$ for all $x\geq 0$. Now using $f$ is odd, we get that $f$ is periodic. If $f$ is constant, then $\boxed{f(x) = 0 \text{ for all }x}$. We'll assume that $f$ is non-constant with period $t (>0)$. We have $f(x^2) = xf(x)$. This gives \[f(xf(x) - x^2 + y^2)=f(f(x^2) - x^2 + y^2)=f(y^2).\]Now $f(c) = f(0) = 0$, so $x = c$ gives $f(-c^2) = f(x)$ for $x\ge 0$. We also get \[f((x+c)^2) = (x+c)f(x)\implies xf(x)=f(x^2)=f(x^2+c^2) = (x+c)f(x)\]if $2x$ is an integer. As $c\neq 0$, $f(x) = 0$ for all $x$ such that $2x$ is integer. Now, $f\left(-\frac{1}{2}\right) - \left(-\frac{1}{2}\right) = \frac{1}{2}$, so we can assume $c = \frac{1}{2}$. Now, $f(c^2) = cf(c) = 0$, so $f\left(-\frac{1}{4}\right)=f\left(\frac{1}{4}\right) = 0$. Therefore we can assume $c = \frac{1}{4}$ also. Therefore $c$ can be $\frac{1}{2^n}$ for any positive integer $n$. Therefore $c$ can be arbitrarily small therefore there is a $c<t$. This is false as $t$ is the period of $f$, so $f$ is constant. $\hfill{\blacksquare}$

TheDarkPrince wrote: Now using $f$ is odd, we get that $f$ is periodic. If $f$ is constant, then $\boxed{f(x) = 0 \text{ for all }x}$. We'll assume that $f$ is non-constant with period $t (>0)$. Can you guarantee that there is such minimal $t$? Functions like Dirichlet functions are periodic, but have no minimal period..

Trolled by that minimum lol

The only solutions are $\boxed{f(x)=x}$ and $\boxed{f(x)=0}$. Let $P(x,y)$ be the assertion $f(f(x)-x+y^2)=yf(y)$ $P(x,0)$ gives $f(f(x)-x)=0 \forall x -(1)$. If $f^{-1}(\{0\})=\{0\}$, then this implies that $f(x)-x=0\Longrightarrow f(x)=x$ $P(1,1) : f(f(1))=f(1)$, so plugging in $x=f(1)$ in $(1)$ gives $f(0)=0$. Note that comparing $P(x,y)$ and $P(x,-y)$ gives $yf(y)=-yf(-y)$, hence $f$ is odd. Now suppose $t(>0)$ is in $ f^{-1}(\{0\})$. $P(t,y)$ yields $f(y^2-t)=yf(y)=f(y^2)$. If $x<0$, we have $f(x)=-f(-x)=-f(-x+t)=f(x-t)$, thus $f$ has period $t$ (and $t^2$ since $f(t^2)=tf(t)=0$). It's easy to see that $f\equiv 0$ is a solution, so now assume that for some $u>0$ $f(u)=s\neq 0$. Combining $P(0,\sqrt{u})$ and $P(u,\sqrt{u})$ gives $$f(u)=\sqrt{u}f(\sqrt{u})=f(f(u))\quad\Longrightarrow\quad f(s)=s$$, and $f(|s|)=|s|$. Let $s_0 =|s|$, then substituting $x=0, y=s_0 +t$ yields $$f(s_{0}^2+2s_0t)=f((s_0+t)^2)=(s_0+t)f(s_0+t)=(s_0+t)f(s_0)=s_0(s_0+t)$$, and by $(1)$, $f(s_0t)=-f(-s_0t)=-f(f(s_0^2+2s_0t)-(s_0^2+2s_0t))=0$. But then $$ts_0+s_0f(s_0)=(s_0+t)f(s_0)=f(s_0^2+2s_0t)=f(s_0^2)=s_0f(s_0)$$, a contradiction since $s_0,t>0$. ($s_0t$ is a period, too!) Hence the only solutions are $\boxed{f(x)=x}$ and $\boxed{f(x)=0}$.

MNJ2357 wrote: $f(s_0^2+2s_0t)=f(s_0^2)$ Why does this work?

this is an easy problem f(x)-x=t f(t+y^2)=yf(y) f(t+y^2)=f(k+y^2)=f(z+y^2)=........ f(t+y^2)=const or t=const f(x)=0 and x

smurfcc wrote: MNJ2357 wrote: $f(s_0^2+2s_0t)=f(s_0^2)$ Why does this work? $f$ has period $s_0t$, (we've shown that $f(s_0t)=0$), so $$f(s_0^2+2s_0t)=f(s_0^2+s_0t)=f(s_0^2)$$

Hypernova wrote: I got $f(x)=x, f(x)=0$ but I'm not confident with my solution How you get $f(0)=0$?

TheDarkPrince wrote: Hopefully this works (it doesn't lol) Hypernova wrote: Find all functions $f$ such that $f:\mathbb{R}\rightarrow \mathbb{R}$ and $f(f(x)-x+y^2)=yf(y)$ Solution. We have $f(f(y^2)) = yf(y)$, so $f(f(1)) = f(1)$. Therefore plugging $x = f(1)$ gives $f(y^2) = yf(y)$. This also means $f(0) = 0$ and replacing $y$ by $-y$ gives $f$ is odd. Now, $\boxed{f(x) = x \text{ for all }x}$ is a valid solution. Therefore we'll now assume that there is some $a$ such that $f(a)\neq a$, so \[f(f(a)-a+y^2) = yf(y) = f(y^2),\]so $f(x+c) = f(x)$ for $x\geq 0$ and some constant $c:=f(a) - a$. Plugging $-a$ gives $f(x-c) = f(x)$ for all $x\geq 0$. Now using $f$ is odd, we get that $f$ is periodic. If $f$ is constant, then $\boxed{f(x) = 0 \text{ for all }x}$. We'll assume that $f$ is non-constant with period $t (>0)$. We have $f(x^2) = xf(x)$. This gives \[f(xf(x) - x^2 + y^2)=f(f(x^2) - x^2 + y^2)=f(y^2).\]Now $f(c) = f(0) = 0$, so $x = c$ gives $f(-c^2) = f(x)$ for $x\ge 0$. We also get \[f((x+c)^2) = (x+c)f(x)\implies xf(x)=f(x^2)=f(x^2+c^2) = (x+c)f(x)\]if $2x$ is an integer. As $c\neq 0$, $f(x) = 0$ for all $x$ such that $2x$ is integer. Now, $f\left(-\frac{1}{2}\right) - \left(-\frac{1}{2}\right) = \frac{1}{2}$, so we can assume $c = \frac{1}{2}$. Now, $f(c^2) = cf(c) = 0$, so $f\left(-\frac{1}{4}\right)=f\left(\frac{1}{4}\right) = 0$. Therefore we can assume $c = \frac{1}{4}$ also. Therefore $c$ can be $\frac{1}{2^n}$ for any positive integer $n$. Therefore $c$ can be arbitrarily small therefore there is a $c<t$. This is false as $t$ is the period of $f$, so $f$ is constant. $\hfill{\blacksquare}$ How you have $f(f(y^2)) = yf(y)$, if you do not know that $f(0)=0$?

@above it's odd function so trivially $f(0)=0$

Hypernova wrote: @above it's odd function so trivially $f(0)=0$ Ok, thank you! In this way, the function it is additive.

Actually i'm pretty sure @2above's reasoning is wrong, cuz the "flip $x$ by $-x$ reasoning fails for $x = 0$. Basically you get $f(f(y^2)) = yf(y)$ from just $y^2, y$, and then he actually gets $f(0) = 0$ later from just plugging in $x = 0$ into $f(x^2) = xf(x)$.

RevolveWithMe101 wrote: Feridimo wrote: this is an easy problem f(x)-x=t f(t+y^2)=yf(y) f(t+y^2)=f(k+y^2)=f(z+y^2)=........ f(t+y^2)=const or t=const f(x)=0 and x LOL Thats totally wrong, dont troll here prove it wrong! If you do not agree with this solution, I will present you the second solution. y=0 f(f(x)-x)=0 then 1)f(x)=0 2)f(x)-x=c....... c=0 f(x)=x If you still disagree with the solution, please show me the error. F(g(x))=const then F(x)=const or g(x)=const

Question for you, @above: what is the range of $f(x)-x$?

Feridimo wrote: If you still disagree with the solution, please show me the error. F(g(x))=const then F(x)=const or g(x)=const This statement is wrong. For example, $g(x)=x^2$ and $f(x)=x-|x|$.

Denote $P(x,y)$ for given assertion. (1) Elementary Part 1. $P(x,-y)$ gives $f(y)=-f(-y)$ if $y \ne 0$. 2. $P(1,1)$ gives $f(f(1))=f(1)$ and $P(f(1), 0)$ gives $f(0)=0$. 3. $P(0,y)$ gives $f(y^2)=yf(y)$ and $P(x,0)$ gives $f(f(x)-x)=0$. If $f(0)$ is the only zero, $f(x)=x$. Else, let $f(a)=0$ for some $a \ne 0$. (2) Proving $f(1/2)=0$ 4. $P(a,y), P(0,y)$ and $P(-a,y)$ gives $f(a+y^2)=f(y^2)$ and $f(-a+y^2)=f(y^2)$. Combining with 1, we get $f(a+z)=f(z)$ for every $z$. 5. For $y=(1/2+a), z=(1/2-a)$, $yf(y)=f(y^2)=f(1/4+a+a^2)=f(1/4-a+a^2)=zf(z)$. Since $y-z=2a$, $f(y)=f(z)=0$. Therefore, $f(1/2)=f(1/2+a)=0$. Also, by 2 and 4, $f(1)=0$. (3) Deducing $f(x)=0$ 6. From 3, $P(y^2,0)$ implies $f(f(y^2)-y^2)=f(yf(y)-y^2)=0$. 7. $P(0, y+1)$ gives $f(y^2 + 2y) = yf(y)+f(y)$ and by 3, $f(yf(y)+f(y)-y^2-2y)=0$. 8. Since $f(f(y)-y)=0$, $f( [yf(y)-f(y^2)]+[f(y)-y]-y)=0$ implies $f(-y)=0$ for all $y$. 9. Therefore $f(y)=0$ if such $a$ exists. By 3, 9, $f(x)=x$ and $f(x)=0$ is the only answer.

Hypernova wrote: Trolled by that minimum lol a.k.a. India Practice TST 2016

MarkBcc168 wrote: Feridimo wrote: If you still disagree with the solution, please show me the error. F(g(x))=const then F(x)=const or g(x)=const This statement is wrong. For example, $g(x)=x^2$ and $f(x)=x-|x|$. If you are talking about real numbers, it is true.but if we look at the numbers that are not true, feridimo is right. x=t*i f(g(x))=-2t!!!

Ferid wrote: If you are talking about real numbers, it is true. And we are. If you want a counterexample over $\mathbb C$, you can find those too, of course. Just consider $g(z) = e^z$ and $f(z) = \begin{cases} 1 & z =0 \\ 0 & z\ne 0 \end{cases}$. Or even $g(x) = |x|^2$ and $f(x) = x-|x|$.

Hypernova wrote: Find all functions $f$ such that $f:\mathbb{R}\rightarrow \mathbb{R}$ and $f(f(x)-x+y^2)=yf(y)$ Didn't read all the anwers submitted but i think this a new solution : denote the given equation as $p(x,y)$ Insert $p(x,0): f(f(x)-x)=0 $ now insert $p(f(x)-x, y): hence f(x-f(x)+y^2)=yf(y) (1) $ Now assume that there exists $x_0$ such that $f(x_0) > 0 $ now insert $ (x_0 , f(x_0)^\frac{1}{2}) in (1) $ and we get$ f( f(x_0)^\frac{1}{2} ) = f(x_0)^\frac{1}{2} $ now insert $ p( f(x_0)^\frac{1}{2} , y) $ and we get $f(y^2)=yf(y) (2)$ hence we have$ f(y^2 + f(x)-x )=f(y^2)$ . Now set x as a constant , and name$ f(x)-x$ as $T$ . Now we have $f(y+T)=f(y) $ for any positive y . Now plug $y+T$ in (2) and set y to a natural number , hence we get : $ f(y^2+2yT+T^2)=f(y^2)=yf(y)=(y+T)f(y+T)= (y+T)f(y) $ and hence we get $Tf(y)=0$ hence T=0, hence $f(x)-x=0$ for any x hence $f(x)=x$ . Now we examine the case in which there are no x such that $f(x)>0$ . Insert a $y<0$ in p(x,y) , now if $f(y)<0 $we get$ f(f(x)-x+y^2) >0$ and contradition. So for every negative y , we must have $f(y)=0$. Now insert $p(-x,-y) for x,y>0 $so we get $f(x)=0$ for positive x. Now insert $p(x, x^\frac{1}{2})$ for positive x and hence we get$ f(0)=0$ , hence we get $f(x)=0$ for any x.$ and we're done.

Let $P(x, y) $ be the assertion : $f(f(x)-x+y^2)=yf(y)$ $P(x,0) \Rightarrow f(f(x) - x) =0.... (1) $ Compare $P(x, y) $ with $p(x,-y) $ we conclude that $f$ is odd, hence $f(0)=0$. $P(0,y) \Rightarrow f(y^2)=yf(y).... (2)$ $ y\rightarrow f(y) $ in $(2) \Rightarrow f(f(y)^2)=f(y)^2 .... (3)$ $P(y^2,y) \Rightarrow f(yf(y)) =yf(y) $ hence $yf(y)$ is a fixed point, and so is $f(f(x) - x+y^2)= f(x-f(x) +y^2)$ (from comparing $p(x, y), p(-x, y))$. Hence $ff(x) =f(x) \forall x\ge 0 $ and since $f$ is odd we get that $ff(x) =f(x) \forall x \in R $. Suppose that there exist $a\neq 0$ such that $f(a) =0$ from $(1)$ we get that $f(a^2)=0$. From $P(a, y) $ we get that $f$ is periodic with period $a$ (and $a^2$). Set $y\rightarrow f(y) +a $ in $(2) \Rightarrow f(f(y)^2+2af(y)) =f(y)^2 + af(y)..... (4)$ Set $x \rightarrow f(y)^2+2af(y) $ in $ (1) \Rightarrow f(af(y)) =0 $ From $(3) $ and $(4)$ we get that : $f(y)^2=f(f(y)^2)=f(y)^2+af(y) $ Hence $af(y)=0 $, so either $ f(y) =0 $ $\forall y\in R $ or $f$ is injective at $0$, hence $(1) $ implies that $f(x) =x $ $\forall x\in R $.

here's a solution without using periodic: easy to observe $f(0)=0$ and $f(y)=-f(-y)$ for nonzero $y$'s. putting $x=-x$ gives us $f(x - f(x) + y^2) = y f(y)$ $*$ let's take a look at equation $y=y^2 - x + f(x)$ ;if there's a solution for that ,for example $g(x)$ (and it does ;according to $*$ ,we can assume that $x > f(x)$),putting $y=g(x)$ gives us $g(x)=1$ and we'll achieve $f(x)=x$ (obviously ,if $y=0$ is only solution we'll achieve this again and if $g(x)=0$ we can easily get $f(x)=0$) and we're done!

Loved this problem a lot! It was quite hard, so my solution is pretty lengthy. The solutions are: $$f(x)=0 \ \ \forall x \in \mathbb{R}$$$$f(x)=x \ \ \forall x \in \mathbb{R}$$ Let $P(x,y)$ denote the original equation. Comparing $P(x,y)$ and $P(x,-y)$, we get that $f$ is odd, except possibly at $0$. Case I: $xf(x) \leq 0$ for all $x$. Note that every sufficiently large $t$ can be written in the form $t=f(x)-x+y^2$ for some $x,y$. Then $P(x,y)$ and our original assumption gives us $f(t) \leq 0$. Hence $f(x)-x$ is not bounded from below, so every real number $t$ can be written in the above mentioned form. Therefore $f(t) \leq 0$ for all real $t$. Hence, from our original assumption, we get $f(x) =0$ for all $x<0$. This combined with oddness gives $f(x)=0$ for all non-zero $x$. Then $P(1,1)$ gives $$f(0)=f(f(1)-1+1^2)=f(1)=0$$So we get the solution $$f(x)=0 \ \ \forall x \in \mathbb{R}$$ Case II: $tf(t)>0$ for some $t$. $P(t^2,t)$ gives $f(f(t^2))=tf(t)$. Now $P(f(t^2),t)$ gives $$tf(t)=f(f(f(t^2))-f(t^2)+t^2) = f(f(-t^2)-(-t^2)+tf(t))$$Let $z= \sqrt{tf(t)}>0$. Now $P(-t^2,z)$ gives $f(z)=z$. Let $$S=\{f(x)-x \mid x \in \mathbb{R} \}$$The above argument shows that $0 \in S$. Claim 1: $u \in S$ $\iff$ $f(u)=0$. Proof: If $u=f(x)-x$ then $P(x,0)$ gives $f(u)=0$. In particular, $f(0)=0$; so $f$ is odd over the entirety of reals. Conversely, if $f(u)=0$ then $f(-u)=0$ by oddness. So we can write $u=f(-u)-(-u)$. $\blacksquare$ Claim 2: $u \in S$ $\implies$ $u$ is a period of $f$. Proof: Comparing $P(u,y)$, $P(0,y)$ and $P(-u,y)$ gives us $f(y^2-u)=f(y^2)=f(y^2+u)$ which combined with oddness gives us the periodicity with period $u$. $\blacksquare$ Claim 3: $S$ forms a group under addition. Proof: We have already proved existence of zero element and additive inverse (the latter follows from oddness of $f$). Let $u,v \in S$ and suppose $u=f(x)-x$. Then by Claim 2, $$u+v= f(x)-(x-v)=f(x-v)-(x-v) \in S$$$\blacksquare$ Claim 4: $xf(x)-x^2 \in S$ for all $x$. Proof: $P(0,x)$ gives $$xf(x)-x^2=f(x^2)-x^2 \in S$$$\blacksquare$ Claim 5: $u \in S$ $\implies$ $u^2 \in S$. Proof: Put $x=-u$ in Claim 4. $\blacksquare$ Claim 6: $f(f(x))=f(x)$ for all $x$. Proof: Comparing $P(0,y)$ and $P(y^2,y)$ gives $f(y^2)=f(f(y^2))$. This, combined with oddness gives the required claim. $\blacksquare$ Claim 7: $f(y)^2-y^2 \in S$ for all $y$. Proof: By Claim 4 and Claim 2, for any real $x$ and $u \in S$, we have $$(x+u)f(x)-(x+u)^2=(x+u)f(x+u)-(x+u)^2 \in S$$$$\implies (xf(x)-x^2) + (-u^2) + u(f(x)-2x) \in S$$Now Claim 5, Claim 4 and Claim 3 give $u(f(x)-2x) \in S$. Taking $x=-f(y)$ in the previous relation and using Claim 6, we get $uf(y) \in S$ for all real $y$ and $u \in S$. In particular $f(y)(f(y)-y) \in S$ for all real $y$. Hence using Claim 3 and Claim 4, we get $$f(y)^2-y^2 = y(f(y)-y)+f(y)(f(y)-y) \in S$$$\blacksquare$ Put $y=x-u$ in Claim 7 for some $u \in S$ and use Claim 2 to get $f(x)^2-(x-u)^2 \in S$ $$\implies (f(x)^2-x^2)+(2xu-u^2) \in S$$Now Claim 3 and Claim 7 give $2xu-u^2 \in S$ for any real $x$ and $u \in S$. If $S$ contains a non-zero number $u$, then the previous expression reaches all real numbers as $x$ ranges over $\mathbb{R}$. Hence by Claim 1 we get $f(x)=0$ for all real $x$, which contradicts our initial assumption. Hence $S=\{ 0 \}$, and so we get the solution $$f(x)=x \ \ \forall x \in \mathbb{R}$$$\blacksquare$ EDIT: After reading some of the above solutions, I realized that Case I and the first paragraph of Case II can be skipped entirely by using $f(f(1))=f(1)$. Oh well.

The only solutions are $f(x) = 0$ and $f(x) = x$, which trivially work. Let $P(x, y)$ be the assertion to the given equation. $P(1, 1)$ gives $f(f(1)) = f(1)$, so $P(f(1), x)$ gives $f(x^2) = xf(x)$. From $x = 0$ we also have $f(0) = 0$ and changing $x$ to $-x$ we also have $f$ is odd. Therefore, we can rewrite $P(x, y)$ as $f(f(x) - x + y^2) = f(y^2)$ and $P(-x, y)$ as $f(f(x) - x - y^2) = f(-y^2)$. Now we have a slightly better equation, $$P(x, y) : f(f(x) - x + y) = f(y) \text{ for all } x, y$$We use the following claim. Claim: Given $x, c \in \mathbb{R}$, $f(x + c) = f(x)$ if and only if $f(c) = 0$. Proof: If $f(x + c) = f(x)$, then comparing $P(x, x+c-f(x))$ and $P(x+c, x+c-f(x))$ gives $f(c) = f(0) = 0$. In the other direction, if $f(c) = 0$, then $f(-c) = 0$, so $P(-c, x)$ gives $f(x+c) = f(x)$. Done. If $f(x) \neq 0$ for all $x \neq 0$, then $P(x, 0)$ gives $f(f(x) - x) = 0$, thus $f(x) = x$ for all $x$, which is a solution. Assume there exists $a \neq 0$ such that $f(a) = 0$. We have $f(a^2) = af(a) = 0$, so by our claim, $$f(1) = f(a^2 + 2a + 1) = (a + 1)f(a + 1) = (a + 1)f(1) \Longrightarrow f(1) = 0$$Now, suppose $b \in f(\mathbb{R})$. Then $P(x, x)$ gives $f(b) = b$ and our claim gives $f(x + 1) = f(x)$, so $$f(b^2 + b + 1) = f(b^2 + b) = f((b+1)f(b)) = f((b+1)f(b+1)) = f(f(b^2 + 2b + 1)) = f(b^2 + 2b + 1)$$and again, our claim gives $f(b) = b = 0$. Hence, $f(x) = 0$ for all $x$, and we are done.

hyay wrote: Therefore, we can rewrite $P(x, y)$ as $f(f(x) - x + y^2) = f(y^2)$ and $P(-x, y)$ as $f(f(x) - x - y^2) = f(-y^2)$. Now we have a slightly better equation, $$P(x, y) : f(f(x) - x + y) = f(y) \text{ for all } x, y$$ I can't understand why $P(x, y) : f(f(x) - x + y) = f(y) \text{ for all } x, y$. From $f(f(x) - x - y^2) = f(-y^2)$, we should say that $P(x, y) : f(f(x) - x + y) = f(y) \text{ for all } x \text{ and } y\le0$.

Rama12 wrote: I can't understand why $P(x, y) : f(f(x) - x + y) = f(y) \text{ for all } x, y$. From $f(f(x) - x - y^2) = f(-y^2)$, we should say that $P(x, y) : f(f(x) - x + y) = f(y) \text{ for all } x \text{ and } y<0$. Yes, it is true that $P(-x, y)$ from the original equation implies $f(f(x) - x + y) = f(y)$ for $y < 0$. The other part, $y \geq 0$, comes from $P(x, y)$ itself. @below Thanks for your kind words!

hyay wrote: Rama12 wrote: I can't understand why $P(x, y) : f(f(x) - x + y) = f(y) \text{ for all } x, y$. From $f(f(x) - x - y^2) = f(-y^2)$, we should say that $P(x, y) : f(f(x) - x + y) = f(y) \text{ for all } x \text{ and } y<0$. Yes, it is true that $P(-x, y)$ from the original equation implies $f(f(x) - x + y) = f(y)$ for $y < 0$. The other part, $y \geq 0$, comes from $P(x, y)$ itself. Oh, I understood. Thanks. I like your solution.

claim: $f(x^2)=xf(x)$ proof: define $g(x)=f(x)-x$ just $P(1,1) $ and $P(f(1),y)$ $\blacksquare$ $P(x,0 ) \implies f(f(x)-x)=0$ now let $f(a)=0 \implies f(a^2)=0$ $P(-a,y) \implies f(y+a)=f(y) : y\ge0$ and similarly $f(y+a^2)=f(y)$ $P(x,y+a)$ for $y \in \mathbb{Z}$ $(y+a)f(y)=yf(y)$ so $a=0$ so $f(x)=x$ or $f(x)=0 \forall x \in Z$ if $f(\mathbb{Z})=0$ put $x=\frac{k}{n}+n$ in $f(x^2)=xf(x)$ we'll have $f(\mathbb{Q})=0$ so $f(x+q)=f(x)$ now put $y=\sqrt{x} \implies f(f(x))=f(x)$ using that $f$ is odd suppose there exists some $a \neq 0 \in f(\mathbb{R})$ so $f(a)=a$ so for $q \in \mathbb{Q}$ $f(a^2+2aq+q^2)=(a+q)f(a+q) \implies f(a^2+2aq)=a^2+aq$ so $f(a^2+aq)=a^2+aq$ since $a^2+aq \in f(\mathbb{R})$ but $f(a^2+aq)=f(a^2+2a\frac{q}{2})=a^2+a\frac{q}{2}=a^2+aq$ implies $a=0$ a contradiction so $f \equiv 0 $

Nice one Let $P(x,y)$ be the following assertion: $$f(f(x)-x+y^2)=yf(y)$$Notice that: $P(1,1) : f(f(1)) = f(1)$ $P(f(1),y) : f(y^2) = yf(y)$ $P(f(1),0) : f(0) = 0$ Easily we get that: $$ -yf(-y) = f(y^2) = yf(y) \implies - f(y) = f(-y) $$Suppose that $t = f(x) - x$, we get: $P(x, y) + P(-x, y) : f(y^2 + t) = yf(y) = f(y^2) = f(y^2 - t)$ WLOG assume $t \geq 0$, and since $f$ is odd then we have that for any $n \in \mathbb{Z}$: $$ f(y+nt) = f(y) \ \ \forall y \in \mathbb{R} $$Hence $f$ is periodic with period $t$, and $$ f(f(x)) = f(x+t) = f(x) $$Notice that if there exitst $t_0 = f(x_0) - x_0 \neq 0$, then: $P(0,y+t_0) : f(y^2 + 2yt_0) = yf(y) + t_0f(y)$ $$ \implies nf(n) = f(n^2 + 2nt_0)= nf(n) + t_0 f(n) \ \ \forall n \in \mathbb{Z} $$$$ \implies f(\mathbb{Z}) = 0 $$Since $1 \in \mathbb{Z}$, then $f$ is periodic with period $t = f(1) - 1 = - 1$. $P(0, f(y) + 1) : f(f(y)^2 + 2f(y)) = f(y)^2 + f(y)$ $P(f(y)^2 + 2f(y), 0) : - f(y) = f(-f(y)) = f(f(f(y)^2 + 2f(y)) - f(y)^2 - 2f(y)) = 0$ Thus if there exists a $x \in \mathbb{R}$ such $f(x) - x \neq 0$ then $f(x) = 0 \ \ \forall x \in \mathbb{R}$, otherwise $f(x) - x = 0 \ \ \forall x \in \mathbb{R}$, and we are done. $\blacksquare$

The answers are \(f\equiv0\) and \(f(x)\equiv x\), which obviously work. Now we will show these are the only solutions. Let \(S=\{f(x)-x:x\in\mathbb R\}\). If \(S\) contains a single element \(k\), then \(f(x)=x+k\) for all \(k\). It is easy to check that the only function of the form \(f(x)\equiv x+k\) that satisfies the functional equation is \(f(x)\equiv x\), i.e.\ only \(k=0\) works, so henceforth assume \(S\) contains at least two distinct elements. Claim: For \(c\in S-S\), we have \(f(x)=f(x+c)\) for all \(x\). Proof. If \(a\in S\) and \(a+c\in S\), observe that \(f(y^2+a)=yf(y)=f(y^2+a+c)\) for all \(y\), so \(f(x)=f(x+c)\) holds whenever \(x\ge a\). However, for large positive integers \(n\), note that \[f(a+nc)-(a+nc)=f(a)-(a+nc)=f(a)-a-nc,\]so we may instead select \(f(a)-a-nc\in S\) and \(f(a)-a-(n-1)c\in S\) (instead of \(a\in S\) and \(a+c\in S\)) to show that \(f(x)=f(x+c)\) holds whenever \(x\ge f(a)-a-nc\). Now take \(n\to\infty\). \(\blacksquare\) Claim: \(\mathbb Z\subseteq S\) Proof. By our asumption that \(|S|\ne1\), we may select \(c\in S-S\) with \(c\ne0\). For \(n\in\mathbb Z\), observe since \(\frac{n+c}2-\frac{n-c}2=c\) and \(\left(\frac{n+c}2\right)^2-\left(\frac{n-c}2\right)^2=nc\) are both integer multiples of \(c\) that \(f\left(\frac{n-c}2\right)=f\left(\frac{n+c}2\right)\) and \begin{align*} \frac{n-c}2\cdot f\left(\frac{n-c}2\right)&=f\left(f(x)-x+\left(\frac{n-c}2\right)^2\right)\\ &=f\left(f(x)-x+\left(\frac{n+c}2\right)^2\right)\\ &=\frac{n+c}2\cdot f\left(\frac{n+c}2\right). \end{align*} It follows that \[f\left(\frac{n-c}2\right)=f\left(\frac{n+c}2\right)=0\]for all integers \(n\). Then \(\frac{n-c}2\) and \(\frac{n+c}2\) are elements of \(S\); in particular, \(1=\frac{2+c}2-\frac c2\in S-S\), so repeat the above argument with \(c=1\) to show \(\frac{n-1}2\) and \(\frac{n+1}2\) are always elements of \(S\). It follows that all integers are in \(S\). \(\blacksquare\) Claim: For all \(x\), we have \(f(x^2)=xf(x)=f(xf(x))\). Proof. Let \(P(x,y)\) denote the assertion. By \(P(x^2,x)\) we have \(f(f(x^2))=xf(x)\). Since \(0\in\mathbb Z\subseteq S\), we may select \(t\) with \(f(t)-t=0\), so \(P(t,x)\) gives \(f(x^2)=xf(x)\). Combining these gives the claim. \(\blacksquare\) Finally, for all \(x\in\mathbb R\), we have \(f(x)-x\in S\), and since \(f(x^2)=xf(x)\), we have \(xf(x)-x^2\in S\). It follows that \begin{align*} f\big(x(x+1)\big)=f\big( (x+1)f(x)\big)=f\big( (x+1)f(x+1)\big)=f\big( (x+1)^2\big), \end{align*}so \(x+1=(x+1)^2-x(x+1)\in S\). It follows that \(S=\mathbb R\), and so \(f\) is constant. Obviously this implies \(f\equiv0\).

Fouad-Almouine wrote: $P(0,y+t_0) : f(y^2 + 2yt_0) = yf(y) + t_0f(y)$ $$ \implies nf(n) = f(n^2 + 2nt_0)= nf(n) + t_0 f(n) \ \ \forall n \in \mathbb{Z} $$$$ \implies f(\mathbb{Z}) = 0 $$ You have $f(y^2+2t_0y+t_0^2)=yf(y) + t_0f(y)$ and $t_0$ is not an integer, so that fails the next equality I think.

VicKmath7 wrote: Fouad-Almouine wrote: $P(0,y+t_0) : f(y^2 + 2yt_0) = yf(y) + t_0f(y)$ $$ \implies nf(n) = f(n^2 + 2nt_0)= nf(n) + t_0 f(n) \ \ \forall n \in \mathbb{Z} $$$$ \implies f(\mathbb{Z}) = 0 $$ You have $f(y^2+2t_0y+t_0^2)=yf(y) + t_0f(y)$ and $t_0$ is not an integer, so that fails the next equality I think. I thought the same~

The only solutions are $\boxed{f\equiv 0}$ and $\boxed{f(x) = x}$, which evidently work. Now we prove they are the only solutions. Let $P(x,y)$ denote the given assertion. $P(1,1): f(f(1)) = f(1)$. $P(f(1), 0)$ gives that $f(0) = 0$. $P(x^2, x): f(f(x^2)) = xf(x)$. $P(0,x): f(x^2) = xf(x)$. Hence \[ f(f(x^2)) = f(x^2) = xf(x)\] Now to see that $f$ is odd, comparing $x$ and $-x$ here gives that $xf(x) = -xf(-x)$, so $f(-x) = -f(x)$ for all $x\ne 0$, and this is clearly true $x = 0$ also as $f(0) = 0$. Define a period of $f$ to be a real number $d$ such that all reals $x$ satisfy $f(x) = f(x+d)$. Claim: For any real number $a$, $f(a) - a$ is a period of $f$. Proof: Suppose otherwise for some $a$. Let $c = f(a) - a$. Now $P(a,y) - P(0,y)$ gives that $f(y^2) = f(y^2 + c)$, so $f(x) = f(x + c)$ for any nonnegative real number $x$. Now, we have $f(-a) - (-a) = -f(a) + a = -c$, so $P(-a,y) - P(0,y)$ gives that $f(x) = f(x - c)$ for any nonnegative real number $x$. For any nonnegative $x$, we have $f(-x + c) = -f(x-c) = -f(x) = f(-x)$, so $f(x) = f(x+c)$ for any real number $x$. $\square$ Claim: If $f(a) = f(b)$ for some $a,b$, then $a - b$ is a period of $f$. Proof: Let $d_1 = f(a) - a, d_2 = f(b) - b$. It suffices to show that $d_2 - d_1$ is a period of $f$. Clearly, $f$ has a period of $d_1$ and $d_2$, so $f(x) = f((x - d_1) + d_1) = f(x - d_1) = f(x + (d_2 - d_1))$. $\square$ Assume $f$ is not the identity. Denote $c > 0$ some period of $f$. For each $x$, notice that $f(f(x) - x - c) = f(f(x) - x) = 0$ (by $P(x,0)$). Claim: For each real number $x$, $cf(x)$ is a period of $f$. Proof: $P(0, f(x) - c): f((f(x) - c)^2) = (f(x) - c) f(x) $ Now, notice that $f(c^2) = cf(c) = 0$, so $c^2$ is a period of $f$. Hence \[f( (f(x) - c)^2) = f(f(x)^2 - 2c f(x) + c^2) = f(f(x)(f(x) - 2c) ) \]Hence $f( f(x) (f(x) - 2c) ) = f(x)(f(x) - c)$, so $f(x) (f(x) - c) - f(x) (f(x) - 2c) = cf(x)$ is a period of $f$. $\square$ $P(0,x + c): f((x+c)^2) = (x+c)f(x+c) = xf(x) + cf(x) = f(x^2) + cf(x)$. Now, we apply $f$ to each side and get that $f(f((x+c)^2)) = f(f(x^2) + cf(x)) = f(f(x^2))$. Since $f(f(y^2)) = f(y^2)$, we get that $f(x^2) = f( (x+c)^2) $, implying that $(x+c)^2 - x^2 = 2xc$ is a period of $f$ for each $x$. Hence every real number is a period of $f$, so $f$ is constant.