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
19.03.2024 20:55
solve this at 1am with bad emotional state, might fakesolve
side note: after failing TST I vowed not to solve an FE ever again, but a nice Ukrainian problem broke my promise.
19.03.2024 21:00
CrazyInMath wrote: solve this at 1am with bad emotional state, might fakesolve
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
19.03.2024 21:02
CrazyInMath wrote: solve this at 1am with bad emotional state, might fakesolve
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*
19.03.2024 21:08
ismayilzadei1387 wrote: CrazyInMath wrote: solve this at 1am with bad emotional state, might fakesolve
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
19.03.2024 23:01
05.07.2024 16:17
CrazyInMath wrote: solve this at 1am with bad emotional state, might fakesolve
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 $0$.
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 $x$.
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 $f(x)=-x$