AoPS - Problem

hakN

05.04.2023 13:04

Let $P(x,y)$ be the assertion. $P(x,0) \implies f(f(0))f(x)=x^2f(0)+f(0)$, if $f(0)\neq 0$ then we have $f(x)=cx^2+d$ for some $c,d\in \mathbb{R}$, plugging back gives $f\equiv 0$, contradicting $f(0)\neq 0$, thus $f(0)=0$. If $f(t)=0$ for some $t\neq 0$, by $P(t,x)$ we have $f(tx)=-t^2f(x)$. By $P(x,t)$ we have $f(x)f(tx)=f(tx) \implies f(x)\in \{0,1\}$ for all $x\in \mathbb{R}$. If $f(x)\neq 0$ for some $x$, then $t^2f(x)+f(tx)=0$ can't hold, thus $f\equiv 0$ in this case. Assume $f\not \equiv 0$, thus we have $f$ is injective at $0$. $P(-\frac{f(x)}{x},x) \implies f(x)^3=-x^2f(-f(x))$. So if $f(a)=f(b)$, we have $a=\pm b$. $P(1,x) \implies f(x+f(x))f(1)=2f(x)$. If $f(a)=f(-a)$ for some $a\neq 0$, then plugging in $x=a$ and $x=-a$ in this gives $f(a+f(a))=f(-a+f(a))$, so either $a=0$ or $f(a)=0$, contradiction. Thus $f$ is injective. $P(-1,x) \implies f(f(x)-x)f(-1)=f(x)+f(-x)$, so we have $f(x)-x=f(-x)+x \implies f(-x)=f(x)-2x$. $P(-x,y) \implies f(f(y)-xy)=\frac{x^2f(y)+f(xy)-2xy}{f(x)-2x}$, plugging $x=\frac{f(y)}{y}$ in this, we have $\frac{f(y)^3}{y^2}+f(f(y))=2f(y)$. Also by $P(-2,-y)$, we have $f(f(y))=\frac{4f(y)-8y+f(2y)}{f(-2)}$. Thus, $f(2y)=8y+(2f(-2)-4)f(y)-f(-2)\frac{f(y)^3}{y^2}$. $P(2,y) \implies f(f(y)+2y)=\frac{8y+2f(-2)f(y)-f(-2)\frac{f(y)^3}{y^2}}{f(2)}$ Putting $y\to -y$, we have $$f(f(y)-4y)=\frac{2f(-2)f(y)-(8+4f(-2))y-f(-2)\frac{(f(y)-2y)^3}{y^2}}{f(2)}$$ Also by $P(-4,y)$, we have $f(f(y)-4y)=\frac{16f(y)+f(-4y)}{f(-4)} = \frac{16f(y)+f(4y)-8y}{f(4)-8}$. Letting $g(y):=\frac{f(y)}{y}$ for $y\neq 0$, we have $f(2y)=y\cdot Q(g(y))$ for some fixed cubic polynomial $Q$. Thus, we have $f(4y)=2y\cdot Q(\frac{Q(g(y))}{2})$. By comparing the two expressions we got for $f(f(y)-4y)$, we get that $R(g(y))=16g(y)+2Q(\frac{Q(g(y))}{2})-8$ for some fixed cubic polynomial $R$. Since degrees of left and right hand sides are $3$ and $9$, respectively, $g(y) = \frac{f(y)}{y}$ can take at most $9$ distinct values. Rewriting the original equation and dividing by $xy$, we have $$x\cdot \left(g(x)g(xy+yg(y))-g(y)\right)=g(xy)-g(x)g(y)g(xy+yg(y)).$$ Thus, for all but finitely many $x$, we have $g(x)g(xy+yg(y))=g(y)$ and $g(xy)=g(x)g(y)g(xy+yg(y))$, which implies $g(y)^2=g(xy)$ for all $y\in \mathbb{R}$. Thus $g(x)=g(1)^2$ for all but finitely many $x$. This gives $g(y)^2 = g(1)^2$ for all $y\in \mathbb{R}$, so $g(y)=g(1)^2$ for all $y\in \mathbb{R}$, implying that either $g\equiv 0$ or $g\equiv 1$. Since $g(x)\neq 0$ for all $x$, we must have $g\equiv 1$ and hence $f(x)=x \ \forall x\in \mathbb{R}$, since $f(0)=0$ as well, we conclude that the only answers are $f(x)=0 \ \forall x\in \mathbb{R}$ and $f(x)=x \ \forall x\in \mathbb{R}$, both of which clearly work.

CANBANKAN

13.04.2023 23:07

The answer is $f(x)\equiv x$ and $f(x)\equiv 0$ only. They clearly work. Let $P(x,y)$ denote the assertion $$f(xy+f(y)) f(x) = f(xy)+x^2f(y)$$$P(x,0)$ yields that $f(f(0))f(x) = (x^2+1)f(0)$. We can check that $f(x)=(x^2+1)c$ works if and only if $c=0$ (say by plugging $x=i$ since this is now a polynomial identity), so we can assume $f(0)=0$. The first key observation is $Q(y)=P(\frac{-f(y)}{y}, y)$ gives $-y^2f(-f(y))=f(y)^3$. Claim: Either $f\equiv 0$ or $f$ is injective. Proof: If $f(y)=f(z)$, $Q(y), Q(z)$ yields $y^2=z^2$. Assume $f(y)=f(-y)$ for some $y\ne 0$. I will first show $f(x)\equiv f(-x)$. By comparing $P(x,y)$ with $P(-x,-y)$ we can see that $(f(x)-f(-x))f(xy+f(y))=0 (*)$ Suppose $xy+f(y)\ne 0, f(xy+f(y))=0$. Let $t=xy+f(y)$. Then $P(t,y)$ yields $f(ty)=-t^2f(y)$, so $f(t^2y)=t^4f(y)$. We can also get $f(t^2)=0$ so $P(t^2,y)$ yields $f(t^2y)=-t^4y$, which implies that $f\equiv 0$. Hence either $f(x)=0 \iff x=0$ or $f(x)\equiv 0$. We suppose the former is true. We can see $f(x)\equiv f(-x)$ (if $xy+f(y)=0$ then plug $-x$ into (*)). Now, $P(x,y), P(-x,y)$ yield $f(-xy-f(y))=f(xy+f(y))=f(-xy+f(y))$ This implies that if $A+B=2f(y), f(A)=f(B)$. In particular, $f(A)=f(-A)=f(2f(y)+A)$. So we say $f(x)=f(x+T)$. Now $P(x,1), P(x+T,1)$ yields $f(1)=0$, contradicting $f(x)=0 \iff x=0$. Henceforth assume $f$ is injective. $$P(-1,y), P(-1,-y) \colon f(-y+f(y)) = f(y)+f(-y) = f(y+f(-y))$$Since $f$ is injective, $-y+f(y)=y+f(-y)$ so $f(y)-f(-y)=2y$. Observation 1: If $f(t)=t$ for some $t\ne 0$, then $f(x)\equiv x$ From $P(x, t)$ and $P(x, -t)$ we can get $f(xt+t)f(x)=f(xt)+x^2t, f(-xt-t)f(x)=f(-xt)-x^2t$. Subtracting the two equations give $2(xt+t)f(x) = 2x^2t + 2xt$, which implies $f(x)\equiv x$ $P(x,1)$ gives $(f(x+C)-1)f(x) = x^2C $ $P(-x-C,1), f(x)-f(-x)=2x$ gives $(f(x+C) - (2x+2C))(f(x)-2x-1) = (x+C)^2C$. This implies that $\lim_{x\to \infty} \frac{f(x)+f(x+C) }{x}=2$ In particular, $f(x)>0$ for all sufficiently large $x$. By AM-GM $C\le 1$. When $C<1$ we note for each chain $x\to x+C \to \cdots$ such that $\lim_{n\to \infty, n \text{ odd }} f(x+nC)/f(x+(n+1)C)$ is equal to a real that is not 1 Basically for some very large real $K$, we've partitioned $\mathbb{R}_{\ge K}$ into two sets $X,Y$, one where $f(x)/x$ tends to $A>1$, one where it tends to $1/A<1$. We can construct $x,y\ge K$ such that $ x\in X, y\in Y$. Asymptotics reveal a contradiction unless $C=1$, in which case we are done by observation 1.

NoctNight

03.05.2023 15:48

this was hard took nearly an hour

The solutions are $f(x)=0$ for all $x$ or $f(x)=x$ for all $x$. These clearly work. Let $P(x,y)$ denote the assertion. Step 1: $f(0)=0$. Proof: $P(0,y)$ gives $f(f(y))f(0)=f(0)$. If $f(0)\neq 0$, then $f(f(y))=0$ for all $y$. $P(x,0)$ gives $$f(f(0))f(x)=x^2f(0)+f(0)$$for all $x$, so $f(f(0))\neq 0$. Thus, $f(x)=ax^2+a$ for some constant $a=\frac{f(0)}{f(f(0))}\neq 0$. Substituting into the original gives LHS degree 6 while RHS degree 4, a contradiction. Step 2: $f$ is injective or constant zero. Proof: Set $xy+f(y)=0$ so $P\left(-\frac{f(y)}{y},y\right)$ gives $$\left(-\frac{f(y)}{y}\right)^2f(y)+f(-f(y))=0$$so $y^2f(-f(y))=-f(y)^3$. Now, if $f(a)=f(b)$ for $a\neq b$ then $a^2=b^2$ or $f(a)=f(b)=0$. In the latter case, WLOG assume $a\neq 0$, then $P(a,1)$ gives $$a^2f(1)+f(a)=0\implies f(1)=0$$Then, $P(1,y)$ gives $2f(y)=0$ so $f(y)=0$ for all $y$. Now assume $f(x)\neq 0$ for $x\neq 0$. If $a^2=b^2$ then $a=-b$ so $f(a)=f(-a)\neq 0$ for $a\neq 0$. Then, $P(a,1)$ and $P(-a,1)$ gives $$f(a+f(1))f(a)=a^2f(1)+f(a)=(-a)^2f(1)+f(-a)=f(-a+f(1))f(-a)$$implying $f(a+f(1))=f(-a+f(1))\implies a+f(1)=\pm(-a+f(1))$. Thus, either $a=0$ or $f(1)=0$, a contradiction as $f(1)=f(0)=0$ but $1\neq \pm0$. Thus, $f$ is injective. Step 3: $f(y)-f(-y)=2y$. Proof: $P(-1,y)$ and $P(-1,-y)$ gives $$f(-y+f(y))f(-1)=f(y)+f(-y)=f(y+f(-y))f(-1)$$so $-y+f(y)=y+f(-y)$ by injectivity ($f(-1)\neq 0$). Thus, $f(y)-f(-y)=2$. Step 4: $f(f(1))=1$. Proof: $P(x,-1)$ gives $$f(-x+f(-1))f(x)=x^2f(-1)+f(-x)$$so since $f(-1)=f(1)-2$ from step 3: $$f(-x+f(1)-2)f(x)=x^2f(-1)+f(-x)=x^2f(1)+f(x)-2x^2-2x$$Substitute $x=\frac{2}{f(1)-2}$ so $x(xf(1)-2x-2)=0$: $$f\left(\frac{-2}{f(1)-2}+f(1)-2\right)=1$$Let this number be $D$, so $f(D)=1$. Now, $P\left(D-f(1),1\right)$ gives $$f(D)f(D-f(1))=(D-f(1))^2f(1)+f(D-f(1))\implies (D-f(1))^2f(1)=0$$so $D=f(1)$ since $f(1)\neq f(0)=0$ by injectivity, as needed. Step 5: $D=f(1)=1$. Proof: $P(D,-1)$ gives $$f(-D+f(-1))=(D-f(-1))^2f(-1)+f(D-f(-1))$$Substituting $D=f(1)$ and $f(-1)=f(1)-2$, we have $f(-2)=4f(-1)+f(2)$ so $$4f(-1)=f(-2)-f(2)=-4\implies f(-1)=1\implies f(1)=f(-1)+2=1$$as desired. Step 6: $f(x)=x$ for all $x$. Proof: We have $f(y)-f(-y)=2y$ and $$f(y)+f(-y)=f(-1)f(f(y)-y)=-f(f(y)-y)$$from step 3. Adding these gives $$2f(y)=2y-f(f(y)-y)=2y-f(y-f(y))+2f(y)-2y=2f(y)-f(y-f(y))$$where the second last equality follows from $f(f(y)-y)=f(y-f(y))-2(y-f(y))$. Thus, $f(y-f(y))=0=f(0)$ for all $y$, so injectivity gives $y-f(y)=0$ so $f(y)=y$ for all real $y$ as desired.

NoctNight

03.05.2023 15:51

what are d-ratings of taiwan tst round 2 quizzes generally?

gghx

04.05.2023 04:36

why so complicated

squarc_rs3v2m

04.05.2023 04:58

bruh beaten by gghx by literal 16 minutes i guess i did $P(x, 0) \implies f^2(0)f(x) = (x^2 + 1)f(0)$; if $f^2(0)\neq 0$ the.. wait nvm gghx just made a typo its the same thing sad ge i guess i also did $P(x, 1)$ implies $f(1) = 0 \implies f(x)^2 = f(x) \implies f(x) \in \{0, 1\}$ for all $x$, then if $f(y) = 1$ then $P(1, y)$ is $1 + f(xy) = 0 \implies f(xy) = -1 \notin \{0, 1\}$ contradiction noctnight ur funny

CANBANKAN

04.05.2023 05:30

Meanwhile I am the biggest loser here. I spent more then 3 hours.

gghx

04.05.2023 06:43

CANBANKAN wrote: Meanwhile I am the biggest loser here. I spent more then 3 hours. skill issue bro do better

IAmTheHazard

16.11.2023 02:52

The answer is $f \equiv 0$ and $f(x)=x$ only, which clearly work. Let $P(x,y)$ denote the assertion. From $P(x,0)$ we find that $f(f(0))f(x)=x^2f(0)+f(0)$, so either $f(0)=0$ or $f$ is (nonlinear) quadratic. Quadratic functions can easily be shown to not work, so it follows that $f(0)=0$. If $f(a)=f(b)$, comparing $P(a,b)$ with $P(b,a)$ implies $a^2=b^2$ or $f(a)=f(b)=0$. On the other hand, if some $c$ has $f(c)=0$ then $P(x,c)$ yields $f(cx)f(x)=f(cx)$, so for any $x$ we either have $f(cx)=0$ or $f(x)=1$. But the latter only occurs at at most two values of $x$, so $f \equiv 0$ on all but (at most) two reals. Repeating this with a different choice of $c$ yields $f \equiv 0$, which is one solution. Otherwise, suppose that $f$ is injective at $0$. I then claim that it's actually injective in general. Indeed, if $f(a)=f(-a) \neq 0$ for some $a \neq 0$, then for any $x \neq 0$ either comparing $P(x,a)$ and $P(-x,-a)$ or $P(x,-a)$ and $P(-x,a)$ yields $f(x)=f(-x)$, since $ax+f(a)=-ax+f(a)=0$ can never hold. But then comparing $P(x,y)$ with $P(-x,y)$ for some $x,y \neq 0$ implies $f(xy+f(y))=f(-xy+f(y))$: a clear contradiction if we take $x=f(y)/y \neq 0$, since now $f(2f(y))=0$ contradicting injectivity at $0$. Now, from $P(-1,y)$ and $P(-1,-y)$ we obtain $f(f(-y)+y)=f(f(y)-y)=f(y)+f(-y)$, so $f(-y)+y=f(y)-y:=k \implies f(y)-f(-y)=2y$ for some $k \in \mathbb{R}$ (dependent on $y$). Solving then yields $f(k)=2k$, but then we also have $f(k)-f(-k)=2k \implies f(-k)=0$, so $k=0$. Thus $f(y)=y$ for all $y$, as desired. $\blacksquare$

mojyla222

24.03.2024 17:02

chengbilly wrote: Find all functions $f : \mathbb{R} \to \mathbb{R}$, such that $$f\left(xy+f(y)\right)f(x)=x^2f(y)+f(xy)$$for all $x,y \in \mathbb{R}$ Proposed by chengbilly Let $P(x,y)$ be the assertion of the problem. One answer is $\boxed{f(x)=0\; \forall x\in \mathbb R}$. So Let's assume that $f$ is not zero constant. Step 1: $f(t)=0 \leftrightarrow t=0$. Define $c:= f(0)$. Then $$P(x,0)\rightarrow cx^2 + c = f(x)f(c) \rightarrow \text{if}\; c\neq 0 \;\text{then}\; f(c)\neq 0 \;\text{and}\; f(x) = \frac{c}{f(c)}(x^2+1)\; \forall x\in \mathbb R,$$which does not work in $P(x,y)$. So $c=0$. Assume that $f(t)=0$ for some $t\neq 0$. $$P(t,y), P(y,t) \rightarrow t^2f(y)+f(ty)=0,\; f(ty) = f(ty)f(y)\rightarrow f(y)\in \{0,1\}$$If $f$ is not constant so $f(y)=1$ for some $y$ and then $f(ty)=-t^2 \notin \{0,1\}$, a contradiction. Step 2: $f$ is injective. Assume that $f(a)=f(b)=s$ for some $a>b$. From last step we infer that $s\neq 0$. $$P(a,b) , P(b,a) \rightarrow sf(ab+s) = a^2s+f(ab) = b^2s+f(ab) \rightarrow a^2=b^2 \rightarrow a=-b.$$So we have that if $f(x)=f(y)\neq 0$ then $x+y=0$. $$P(a,1) , P(-a,1) \rightarrow a^2f(1) + s = sf(a+f(1)) = sf(-a+f(1)) \rightarrow f(a+f(1)) = f(-a+f(1)).$$So $a+f(1) = -a+f(1)=0$ or $a+f(1)-a+f(1)=0$ which imply $f(1)=0\; \text{or}\; a=0$, contradiction. Step 3: $f(-y) = f(y)-2y$ for all $y\in \mathbb R$. $$P(-1,y), P(-1,-y) \rightarrow f(y)+f(-y) = f(-1)f(f(y)-y) = f(-1)f(f(-y)+y) \rightarrow f(y)-y = f(-y)+y \rightarrow f(-y) = f(y)-2y. $$ Step 4: $f(1)=1$ and $f(-1)=-1$. Define $a: = f(1)$ and $b:= f(-1)$. from previous step we know $a=b+2$. $$P(1,1) \rightarrow f(a+1) = 2 \rightarrow f(-a-1) = -2a.$$Then we have $$P(-a-1,1)\rightarrow -2ab= -2a+a(a+1)^2 \rightarrow (a-1)(a+4)=0.$$In addition from $P(1,-1-a)$ we know $f(-3a-1)=-4$. So if $f(1)=a=-4$ then $f(-3a-1) = f(11)=f(1)=-4$ which is contradiction because $f$ is injective. So $a=1$. Step 5: $f(x) =x$ for all real $x\neq \frac{1}{2}$. $$P(x,1) \rightarrow f(x+1) = 1+\dfrac{x^2}{f(x)} \rightarrow f(x)-1 = \dfrac{(x-1)^2}{f(x-1)} = \dfrac{(x-1)^2}{2x-2+f(1-x)}$$On the other hand $$P(-x,1) \rightarrow f(1-x) = 1+\dfrac{x^2}{f(-x)} = 1+\dfrac{x^2}{f(x)-2x}.$$So $$\dfrac{(x-1)^2}{f(x)-1} - 2(x-1) = 1 + \dfrac{x^2}{f(x)-2x}\rightarrow f(x)^2(2x-1) - f(x)\left(4x^2-2x\right)+(2x^3-x^2) =0$$$$\rightarrow (2x-1)(f(x)-x)^2 =0. \rightarrow f(x) =x \; \forall x\neq \frac{1}{2}.$$ Step 6: $\boxed{f(x) =x \; \forall x \in \mathbb R}$. It is just enough to prove $f(\frac{1}{2})=\frac{1}{2}$. We have $$P(-\frac{1}{2},1) \rightarrow f(\frac{1}{2}) = 1+ \dfrac{(-\frac{1}{2})^2}{f(-\frac{1}{2})} = \frac{1}{2}.$$