Find all functions $f:\mathbb{R} \to \mathbb{R}$, such that for any $x, y \in \mathbb{R}$ holds the following: $$f(x)f(yf(x)) + yf(xy) = xf(xy) + y^2f(x)$$ Proposed by Mykhailo Shtandenko
Problem
Source: Ukrainian Mathematical Olympiad 2024. Day 1, Problem 10.4
Tags: function, algebra, functional equation
CrazyInMath
19.03.2024 20:55
solve this at 1am with bad emotional state, might fakesolve
$P(0,0)\Longrightarrow f(0)=0$
$P(1,1)\Longrightarrow f(1)f(f(1))=f(1)$.
If $f(1)=0$ then $P(1,y)$ gives $yf(y)=f(y)$ so $f(x)=0$ for all $x$.
Now let's assume $f(f(1))=1$
$P(x,1)\Longrightarrow f(x)f(f(x))=xf(x)$. So we have $f(f(x))=x$ if $f(x)\neq 0$.
However this works for $f(x)=0$ as well, so we have $f(f(x))=x$ and $f$ is bijective.
$P(x,x)\Longrightarrow f(x)f(xf(x))=x^2f(x)\Longrightarrow f(xf(x))=x^2$.
substitute $x$ with $-x$ we have $f(xf(x))=x^2=f(-xf(-x))$, with injectivity we have $xf(x)=-xf(-x)$ so $f$ is odd function.
$P(x,f(x))\Longrightarrow f(x)f(f(x)^2)+f(x)f(xf(x))=xf(xf(x))+f(x)^3$.
With $f(xf(x))=x^2$ it become $f(x)f(f(x)^2)+x^2f(x)=x^3+f(x)^3$.
Plug in $x=f(1)$ gives $f(1)+f(1)^2=f(1)^3+1$, solving gives $f(1)=\pm1$.
If $f(1)=1$ then $P(1,y)$ gives $f(y)+yf(y)=f(y)+y^2$, so $f(y)=y$ for all $y$.
If $f(1)=-1$ the $P(1,y)$ with $f$ odd gives $-f(-y)+yf(y)=f(y)+yf(y)=f(y)-y^2$, so $f(y)=-y$ for all $y$
The solutions are $f(x)=0$ and $f(x)=x$ and $f(x)=-x$, and they all work.
side note: after failing TST I vowed not to solve an FE ever again, but a nice Ukrainian problem broke my promise.
ismayilzadei1387
19.03.2024 21:00
CrazyInMath wrote: solve this at 1am with bad emotional state, might fakesolve
$P(0,0)\Longrightarrow f(0)=0$
$P(1,1)\Longrightarrow f(1)f(f(1))=f(1)$.
If $f(1)=0$ then $P(1,y)$ gives $yf(y)=f(y)$ so $f(x)=0$ for all $x$.
Now let's assume $f(f(1))=1$
$P(x,1)\Longrightarrow f(x)f(f(x))=xf(x)$. So we have $f(f(x))=x$ if $f(x)\neq 0$.
However this works for $f(x)=0$ as well, so we have $f(f(x))=x$ and $f$ is bijective.
$P(x,x)\Longrightarrow f(x)f(xf(x))=x^2f(x)\Longrightarrow f(xf(x))=x^2$.
substitute $x$ with $-x$ we have $f(xf(x))=x^2=f(-xf(-x))$, with injectivity we have $xf(x)=-xf(-x)$ so $f$ is odd function.
$P(x,f(x))\Longrightarrow f(x)f(f(x)^2)+f(x)f(xf(x))=xf(xf(x))+f(x)^3$.
With $f(xf(x))=x^2$ it become $f(x)f(f(x)^2)+x^2f(x)=x^3+f(x)^3$.
Plug in $x=f(1)$ gives $f(1)+f(1)^2=f(1)^3+1$, solving gives $f(1)=\pm1$.
If $f(1)=1$ then $P(1,y)$ gives $f(y)+yf(y)=f(y)+y^2$, so $f(y)=y$ for all $y$.
If $f(1)=-1$ the $P(1,y)$ with $f$ odd gives $-f(-y)+yf(y)=f(y)+yf(y)=f(y)-y^2$, so $f(y)=-y$ for all $y$
The solutions are $f(x)=0$ and $f(x)=x$ and $f(x)=-x$, and they all work.
side note: after failing TST I vowed not to solve an FE ever again, but a nice Ukrainian problem broke my promise. which country did you get your TST? btw your solution seems correct
Geometrystormer
19.03.2024 21:02
CrazyInMath wrote: solve this at 1am with bad emotional state, might fakesolve
$P(0,0)\Longrightarrow f(0)=0$
$P(1,1)\Longrightarrow f(1)f(f(1))=f(1)$.
If $f(1)=0$ then $P(1,y)$ gives $yf(y)=f(y)$ so $f(x)=0$ for all $x$.
Now let's assume $f(f(1))=1$
$P(x,1)\Longrightarrow f(x)f(f(x))=xf(x)$. So we have $f(f(x))=x$ if $f(x)\neq 0$.
However this works for $f(x)=0$ as well, so we have $f(f(x))=x$ and $f$ is bijective.
$P(x,x)\Longrightarrow f(x)f(xf(x))=x^2f(x)\Longrightarrow f(xf(x))=x^2$.
substitute $x$ with $-x$ we have $f(xf(x))=x^2=f(-xf(-x))$, with injectivity we have $xf(x)=-xf(-x)$ so $f$ is odd function.
$P(x,f(x))\Longrightarrow f(x)f(f(x)^2)+f(x)f(xf(x))=xf(xf(x))+f(x)^3$.
With $f(xf(x))=x^2$ it become $f(x)f(f(x)^2)+x^2f(x)=x^3+f(x)^3$.
Plug in $x=f(1)$ gives $f(1)+f(1)^2=f(1)^3+1$, solving gives $f(1)=\pm1$.
If $f(1)=1$ then $P(1,y)$ gives $f(y)+yf(y)=f(y)+y^2$, so $f(y)=y$ for all $y$.
If $f(1)=-1$ the $P(1,y)$ with $f$ odd gives $-f(-y)+yf(y)=f(y)+yf(y)=f(y)-y^2$, so $f(y)=-y$ for all $y$
The solutions are $f(x)=0$ and $f(x)=x$ and $f(x)=-x$, and they all work.
side note: after failing TST I vowed not to solve an FE ever again, but a nice Ukrainian problem broke my promise. I can definitely agree with the quote *nice Ukrainian problem*
CrazyInMath
19.03.2024 21:08
ismayilzadei1387 wrote: CrazyInMath wrote: solve this at 1am with bad emotional state, might fakesolve
$P(0,0)\Longrightarrow f(0)=0$
$P(1,1)\Longrightarrow f(1)f(f(1))=f(1)$.
If $f(1)=0$ then $P(1,y)$ gives $yf(y)=f(y)$ so $f(x)=0$ for all $x$.
Now let's assume $f(f(1))=1$
$P(x,1)\Longrightarrow f(x)f(f(x))=xf(x)$. So we have $f(f(x))=x$ if $f(x)\neq 0$.
However this works for $f(x)=0$ as well, so we have $f(f(x))=x$ and $f$ is bijective.
$P(x,x)\Longrightarrow f(x)f(xf(x))=x^2f(x)\Longrightarrow f(xf(x))=x^2$.
substitute $x$ with $-x$ we have $f(xf(x))=x^2=f(-xf(-x))$, with injectivity we have $xf(x)=-xf(-x)$ so $f$ is odd function.
$P(x,f(x))\Longrightarrow f(x)f(f(x)^2)+f(x)f(xf(x))=xf(xf(x))+f(x)^3$.
With $f(xf(x))=x^2$ it become $f(x)f(f(x)^2)+x^2f(x)=x^3+f(x)^3$.
Plug in $x=f(1)$ gives $f(1)+f(1)^2=f(1)^3+1$, solving gives $f(1)=\pm1$.
If $f(1)=1$ then $P(1,y)$ gives $f(y)+yf(y)=f(y)+y^2$, so $f(y)=y$ for all $y$.
If $f(1)=-1$ the $P(1,y)$ with $f$ odd gives $-f(-y)+yf(y)=f(y)+yf(y)=f(y)-y^2$, so $f(y)=-y$ for all $y$
The solutions are $f(x)=0$ and $f(x)=x$ and $f(x)=-x$, and they all work.
side note: after failing TST I vowed not to solve an FE ever again, but a nice Ukrainian problem broke my promise. which country did you get your TST? btw your solution seems correct I'm from Taiwan
VicKmath7
19.03.2024 23:01
The solutions are $f(x)=0$, $f(x)=x$ and $f(x)=-x$, which are easily seen to work.
Suppose that $f$ is non-constant. Setting $x=y=0$ gives $f(0)=0$. We claim that $f$ is injective at $0$. Suppose that $f(c)=0$ for $c \neq 0$ and plugging in $x=c$ yields $(c-y)f(cy)=0$, so $f(x)=0$ for all $x \neq c^2$. Choose some real $k$, such that $k \neq \frac{c^2}{k}, 1$ and setting $x=k, y=\frac{c^2}{k}$ yields $f(c^2)=0$, so $f(x)=0$ for all $x$, contradiction.
Plugging in $y=1$ yields $f(x)f(f(x))=xf(x)$, so $f(f(x))=x$ if $f(x) \neq 0$ (i.e. when $x \neq 0$). But $f(f(0))=0$, so $f(f(x))=x$ for all $x$. Setting $x=y$ gives $f(x)f(xf(x))=x^2f(x)$, so similarly we obtain $f(xf(x))=x^2$ for all $x$. Setting $x:=f(x)$ in the above relation and combining with $f(f(x))=x$ we obtain $x^2=f(xf(x))=f(x)^2$, so $f(x) \in \{x, -x\}$ for all $x$.
Suppose for the sake of obtaining contradiction that $f(a)=a$, $f(b)=-b$ for some non-zero $a \neq b$. Setting $x=a, y=b$ yields $af(ab)+bf(ab)=af(ab)+b^2f(a)$, so $f(ab)=ab$. Plugging $x=b, y=a$ gives $-bf(-ab)+af(ab)=bf(ab)-a^2b=ab^2-a^2b$, so $f(-ab)=2a^2-ab \neq ab, -ab$, contradiction. Hence $f(x)=x$ for all $x$, $f(x)=-x$ for all $x$, or $f$ is constant, i.e. $f \equiv 0$, as claimed.
OptimalFEian
05.07.2024 16:17
CrazyInMath wrote: solve this at 1am with bad emotional state, might fakesolve
$P(0,0)\Longrightarrow f(0)=0$
$P(1,1)\Longrightarrow f(1)f(f(1))=f(1)$.
If $f(1)=0$ then $P(1,y)$ gives $yf(y)=f(y)$ so $f(x)=0$ for all $x$.
Now let's assume $f(f(1))=1$
$P(x,1)\Longrightarrow f(x)f(f(x))=xf(x)$. So we have $f(f(x))=x$ if $f(x)\neq 0$.
However this works for $f(x)=0$ as well, so we have $f(f(x))=x$ and $f$ is bijective.
$P(x,x)\Longrightarrow f(x)f(xf(x))=x^2f(x)\Longrightarrow f(xf(x))=x^2$.
substitute $x$ with $-x$ we have $f(xf(x))=x^2=f(-xf(-x))$, with injectivity we have $xf(x)=-xf(-x)$ so $f$ is odd function.
$P(x,f(x))\Longrightarrow f(x)f(f(x)^2)+f(x)f(xf(x))=xf(xf(x))+f(x)^3$.
With $f(xf(x))=x^2$ it become $f(x)f(f(x)^2)+x^2f(x)=x^3+f(x)^3$.
Plug in $x=f(1)$ gives $f(1)+f(1)^2=f(1)^3+1$, solving gives $f(1)=\pm1$.
If $f(1)=1$ then $P(1,y)$ gives $f(y)+yf(y)=f(y)+y^2$, so $f(y)=y$ for all $y$.
If $f(1)=-1$ the $P(1,y)$ with $f$ odd gives $-f(-y)+yf(y)=f(y)+yf(y)=f(y)-y^2$, so $f(y)=-y$ for all $y$
The solutions are $f(x)=0$ and $f(x)=x$ and $f(x)=-x$, and they all work.
side note: after failing TST I vowed not to solve an FE ever again, but a nice Ukrainian problem broke my promise. I have doubts about the statement "However, this works for f(x) = 0 as well". This means that if $f(x) = 0$, then $x = f(f(x)) = f(0) = 0$. However, it has not been shown at that point that $f$ is injective at $0$.
megarnie
02.08.2024 18:00
The only solutions are $f(x) = x$, $f(x) = -x$, and $f\equiv 0$. These work. Now we prove they are the only ones. Let $P(x,y)$ denote the given assertion. Clearly $0$ is the only constant solution, so assume $f$ isn't constant. $P(0,0): f(0) = 0$. $P(x,1): f(x) (f(f(x)) + 1) = f(x) (x + 1)$. Thus, either $f(x) = 0$ or $f(f(x)) = x$. $P(1, x): f(1) f(x f(1)) + xf(x) = f(x) + x^2 f(1)$. If $f(1) = 0$, then $xf(x) = f(x)\implies f(x) (x - 1) = 0$, so $f(x) = 0\forall x\ne 1$, meaning $f$ is constant, absurd. Hence $f(1) \ne 0$. Claim: If $f(k) = 0$ for some $k \ne 0$, then $f(k^2) \ne 0$ and $f(x) = 0$ for all reals $x\ne k^2$. Proof: Setting $x = k$ in the equation gives $y f(ky) = k f(ky)$, so if $f(ky) \ne 0$, then $y = k$. Thus, for $y \ne k$, we have $f(ky) = 0$, so $f(x) = 0$ for all reals $x\ne k^2$. Since $f$ isn't constant, we have $f(k^2) \ne 0$. $\square$ Claim: $f$ is injective at $0$. Proof: Suppose there existed $k\ne 0$ with $f(k) = 0$. By our claim, we have $f(x) = 0$ for all reals $x\ne k^2$. However, $f(1) = 0$, so $k^2 = 1$, implying that $f(-1) = 0$. However, now note that $f(1434) = 0$ since $1434 \ne (-1)^2$, so $f(1434^2) \ne 0$, absurd as $1434^2 \ne (-1)^2$. $\square$ Thus, if $f(x) = 0$, then $x = 0$. Now since $f(x) = 0$ or $f(f(x)) = x$, we have $f(f(x)) = x$ for all $x\ne 0$, but it is also true for $x= 0$, so $f$ is an involution, and thus bijective. $P(f(x), y): x f(xy) + y f(y f(x) ) = f(x) f(y f(x)) + x y^2$. $P(x,x): f(x) f(xf(x)) = x^2 f(x)$, so $f(xf(x)) = x^2$ for $x\ne 0$, but this is true for all reals $x$ since it's true for $x = 0$ also. Comparing $x$ and $f(x)$ here gives that $f(x)^2 = x^2$, so $f(x) \in \{-x, x\}$ for all reals $x$. Claim: $f(x) = -f(-x)$ for all reals $x$. Proof: This is clearly true for $x = 0$, so assume $x\ne 0$. Suppose we had $f(x) \ne - f(-x)$ for some $x \ne 0$. Then if $f(x) = x$, then $f(-x) = x$ also, but this contradicts injectivity. If $f(x) = -x$, then $f(-x) = x$ since $f(f(x)) =x$. $\square$ Case 1: $f(1) = 1$ $P(1, x): f(x) + xf(x) = f(x) + x^2$, so $xf(x) = x^2$, meaning $f(x) = x\forall x\ne 0$. This is also true for $x= 0$, so $f(x) = x$ for all reals $x$. Case 2: $f(1) = -1$ $P(1,x): -f(-x) + xf(x) = f(x) - x^2$, so $xf(x) = -x^2$, meaning $f(x) = -x$ for all $x\ne 0$, but this is clearly true for $x= 0$. Thus, $f(x) = -x$ for all reals $x$.
A-_AR2_-A
03.08.2024 00:20
MS_Kekas wrote: Find all functions $f:\mathbb{R} \to \mathbb{R}$, such that for any $x, y \in \mathbb{R}$ holds the following: $$f(x)f(yf(x)) + yf(xy) = xf(xy) + y^2f(x)$$ Proposed by Mykhailo Shtandenko Nice problem and even better solution For constant $f$ only solution is clearly $f(x)=0$ Now for other solutions $P(0,0) \implies f(0)=0$ $P(x,1) \implies f(x)ff(x)=xf(x)\implies f$ is injective (easy to see otherwise $f=0$ discussed) so for $x\neq0 \implies ff(x)=x$ (and this is also true for x=0) So $f$ is involutive so $\implies$ $f$ is bijective $P(x,x) \implies f(x)f(xf(x))=x^2f(x) \implies f(xf(x))=x^2$ ..…..(1)……. for all real $x\neq0$ We substitute in ……(1)…… $x \,\to\,f(x)$ we get $f(f(x)ff(x))=f(f(x)x)=(f(x))^2=x^2 \implies f(x)=\pm x$ So there $\exists a,b\neq0$ such that $f(a)=a$ and $f(b)=-b$ $P(a,x) \implies af(ax)+xf(ax)=af(ax)+ax^2 \implies f(ax)=ax$ now we substitute $x\,\to\,\frac{b}{a} \implies f(b)=b=-b$ contradiction And so we deduce all solutions $f(x)=0$ and $f(x)=x$ and $f(x)=-x$