Functional equation wrapped in f's

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Functional equation wrapped in f's

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Source: RMM 2019 Problem 5

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62861

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62861

#1 h Feb 24, 2019, 12:01 PM • 8 Y

Y by Mathuzb, thepiercingarrow, Euler_88, megarnie, aopsuser305, tiendung2006, Adventure10, ngduchieu1903

Determine all functions $f: \mathbb{R} \to \mathbb{R}$ satisfying
$f(x + yf(x)) + f(xy) = f(x) + f(2019y),$ for all real numbers $x$ and $y$ .

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rmtf1111

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rmtf1111

#2 h Feb 24, 2019, 12:02 PM • 10 Y

Y by AlastorMoody, MK4J, ywq233, Pluto1708, mijail, megarnie, Adventure10, khanh20, jannatiar, raspa.abol

-Knock knock!
-Who's there?
-Iran 2016
Solution

This post has been edited 2 times. Last edited by rmtf1111, Feb 24, 2019, 2:32 PM
Reason: @below, thanks

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ubermensch

820 posts

ubermensch

#4 h Feb 24, 2019, 2:20 PM • 4 Y

Y by Djaliloo, Adventure10, Mango247, alexanderhamilton124

rmtf1111 wrote:

-Knock knock!
-Who's there?
-Iran 2016
Solution

Actually $f\equiv c$ for any constant $c$ works. Just put it in the equation and check.

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stroller

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stroller

#6 h Feb 25, 2019, 2:57 PM • 1 Y

Y by Adventure10

rmtf1111 wrote:

We distinguish three cases - either $A=\{0\}$ or $B=\{1\}$ or $f$ is trivially constant.

How do we get this line from the previous argument?

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rmtf1111

698 posts

rmtf1111

#7 h Feb 25, 2019, 3:01 PM • 2 Y

Y by Adventure10, Mango247

stroller wrote:

rmtf1111 wrote:

We distinguish three cases - either $A=\{0\}$ or $B=\{1\}$ or $f$ is trivially constant.

How do we get this line from the previous argument?

If neither of $A$ or $B$ is $\{0\}$ or $\{1\}$ respectively, then there exists a $x\in A$ and $m \in B$ such that $x(m-1)\neq 0$ and so by varying $z$ the expression $y+x(m-1)z$ will span the whole real axis, making $f$ constant

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lminsl

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lminsl

#8 h Feb 26, 2019, 1:04 PM • 11 Y

Y by Morskow, Hjlee, A_Math_Lover, BobaFett101, nguyenhaan2209, Loppukilpailija, Aryan-23, hakN, FateGrandOrder, Adventure10, Mango247

Solution

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william122

1576 posts

william122

#10 h Feb 4, 2020, 3:44 PM • 2 Y

Y by thepiercingarrow, Adventure10

I claim that the set of solutions are $f(x)\equiv c$ , $f(x)=\begin{cases}0& x\neq 0\\ c& x=0\end{cases}$ , and $f(x)=2019-x$ .

If we denote the assertion as $P(x,y)$ , note that $P(2019,y)$ yields $f(2019+yf(2019))=f(2019)$ . If $f(2019)\neq 0$ , we can modify $y$ such that $2019+yf(2019)$ ranges over all reals. Hence, $f\equiv c$ for some constant $c$ .

If $f(2019)=0$ , note that $P(x,1)$ gives $f(x+f(x))=0$ . Now, suppose we have a $z\neq 0$ such that $f(z)=0$ . (note that if $z=0$ , then $P(0,y)$ would yield $f$ is constant). $P(z,x)$ yields that $f(zy)=f(2019y)$ , or equivalently $f(x)=f\left(\frac{2019}{z} x\right)$ . Now, consider $P\left(x_0,\frac{c}{f(x_0)}\right)$ and $P\left(\frac{2019x_0}{z},\frac{c}{f(x_0)}\right)$ for some $x_0$ such that $x_0,f(x_0)\neq 0$ (If such an $x_0$ does not exist, then we will get our second set of solutions, which do indeed work.) They have the same RHS, and comparing LHS gives that $f\left(\frac{2019x_0}{z}+c\right)=f(x_0+c)$ . As $c$ is arbitrary, we get that $f$ has period $\frac{2019x_0}{z}-x_0=p$ .

Now, we consider $P(x,y)$ and $P(x+p,y)$ , which gives that $f(xy)=f(xy+xp)$ . Varying $xy$ and $x$ , we see that if $p\neq 0$ , we get that $f$ is constant. So, we must have $p=0\implies z=2019$ . So, $2019$ is our only root.

However, looking back at $P(x,1)$ , we see that $x+f(x)=2019$ . So, we must have $f(x)=2019-x$ , which indeed satisfies the assertion. Thus, we see that the three aforementioned solution forms are the only ones.

This post has been edited 1 time. Last edited by william122, Feb 5, 2020, 3:48 AM

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niyu

830 posts

niyu

#12 h Feb 9, 2020, 3:11 AM • 3 Y

Y by SerdarBozdag, Adventure10, Seungjun_Lee

We claim that the answers are $f(x) = \begin{cases} 0 & \mbox{if } x \neq 0 \\ c & \mbox{if } x = 0 \end{cases}$ , $f(x) = c$ , $f(x) = 2019 - x$ , where $c \in \mathbb{R}$ is a constant. It is easy to check that these functions work. We now show these are all.

Let $P(x, y)$ denote the given assertion. From $P(2019, y)$ , we get
$\begin{align*} f(2019 + yf(2019)) + f(2019y) &= f(2019) + f(2019y) \\ f(2019 + yf(2019)) &= f(2019). \end{align*}$ If $f(2019)$ is nonzero, as $y$ varies, $2019 + yf(2019)$ achieves all real values, implying that $f(x) = f(2019)$ for all $x$ . Thus, in this case, $f$ is constant. Henceforth, suppose $f(2019) = 0$ , and that $f$ is not identically $0$ .

Now, suppose there exists $u \neq 2019$ for which $f(u) = 0$ . From $P(u, y)$ , we have
$\begin{align*} f(u + yf(u)) + f(uy) &= f(u) + f(2019y) \\ f(uy) &= f(2019y). \end{align*}$ Thus, $f(uy) = f(2019y)$ for all $y$ . Now, from $P(ux, y)$ and $P(2019x, y)$ , we respectively have
$\begin{align*} f(ux + yf(ux)) + f(uxy) &= f(ux) + f(2019y) \\ f(2019x + yf(2019x)) + f(2019xy) &= f(2019x) + f(2019y). \end{align*}$ Since $f(uy) = f(2019y)$ for all $y$ , we find that $f(ux + yf(ux)) = f(2019x + yf(2019x))$ .
Additionally, since $f$ is not identically $0$ , there exists $t$ for which $f(2019t) \neq 0$ . Now, we have
$\begin{align*} f(ut + yf(ut)) &= f(2019t + yf(2019t)) \\ f(ut + yf(ut)) &= f(ut + yf(ut) + (2019 - u)t). \end{align*}$ Since $f(ut) = f(2019t) \neq 0$ , $ut + yf(ut)$ achieves all real values as $y$ varies across $\mathbb{R}$ . Thus, $f(x) = f(x + (2019 - u)t)$ for all $x$ , implying that $f$ is periodic with period $(2019 - u)t$ .

Now, suppose $t \neq 0$ . Let $k = (2019 - u)t$ (which is nonzero because $u \neq 2019$ ). From $P(x + k, y)$ , we have
$\begin{align*} f(x + k + yf(x + k)) + f(xy + ky) &= f(x + k) + f(2019y). \end{align*}$ Noting that $f(x + k) = f(x)$ , we have
$\begin{align*} f(x + yf(x)) + f(xy + ky) &= f(x) + f(2019y). \end{align*}$ Comparing with $P(x, y)$ yields
$\begin{align*} f(xy + ky) &= f(xy). \end{align*}$ Now, fix $y \neq 0$ . As $x$ varies across the real numbers, $xy$ achieves all real values; thus, we have $f(z + ky) = f(z)$ for all $z \in \mathbb{R}$ , implying that $ky$ is also a period of $f$ . However, since $k \neq 0$ , as $y$ varies across the real numbers, $ky$ achieves all real numbers. This is enough to imply that every real number is a period of $f$ , which is enough to imply that $f$ is constant. Thus, $f(x) = f(2019) = 0$ for all $x$ ; however, we had assumed that $f$ is not identically zero, contradiction. Thus, we must have $t = 0$ , meaning that $f(x) = \begin{cases} 0 & \mbox{if } x \neq 0 \\ c & \mbox{if } x = 0 \end{cases}$ , where $c \neq 0$ is a real constant.

Otherwise, $u$ does not exist, meaning that $f(r) = 0$ only for $r = 2019$ . Then, from $P(x, 1)$ , we have
$\begin{align*} f(x + f(x)) + f(x) &= f(x) + f(2019) \\ f(x + f(x)) &= 0. \end{align*}$ This yields $x + f(x) = 2019$ , implying that $f(x) = 2019 - x$ .

Thus, the only solutions to the given functional equation are $f(x) = c$ (where $c \in \mathbb{R}$ is a constant) and $f(x) = 2019 - x$ , as claimed. $\Box$

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TheUltimate123

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TheUltimate123

#13 h Feb 27, 2020, 8:26 AM • 2 Y

Y by rashah76, Pluto1708

Let $P(x,y)$ denote the assertion. The answer is

$f$ constant;
$f(x)=0$ for $x\ne0$ ; and
$f(x)=2019-x$ for all $x$ .

These can be easily verified. We now check these are the only answers. Assume that $f$ is nonconstant. If $f(0)=0$ , then $P(0,y)$ gives $f(2019y)=0$ , so $f\equiv0$ ; thus we may assume $f(0)\ne0$ .

First, $P(2019,y)$ gives $f(2019+yf(2019))=f(2019)$ , so if $f(2019)\ne0$ , then $2019+yf(2019)$ is surjective and $f$ is constant. Henceforth $f(2019)=0$ . Now $P(x,1)$ and $P(0,1)$ give $f(x+f(x))=f(f(0))=f(2019).$ If there is no $z\ne2019$ with $f(z)=2019$ , then $x+f(x)=2019$ for all $x$ , so $f(x)=2019-x$ .

Henceforth assume such $z$ exists, and let $a=z/2019\ne1$ . Assertions $P(2019x,y)$ and $P(zx,y)$ give $f(2019x+yf(2019x))=f(zx+yf(zx))=f(zx+yf(2019x)).$ If $x_0=2019x$ , then we have $f(x_0+yf(x_0))=f(ax_0+yf(x_0))$ . Assume we can select $x_0$ such that $x_0\ne0$ and $f(x_0)\ne0$ ; otherwise we are done. From this, $x_0+yf(x_0)$ attains all real numbers, so we have $f(x)=f(x+(a-1)x_0)$ for all $x$ .

Denote $p=(a-1)x_0$ , so that $f$ is periodic modulo $p$ . To finish, note that $P(x,y)$ and $P(x+p,y)$ yield $f(xy+py)=f(xy)$ . As we fix $xy$ and vary $y$ , we have $f$ is constant, so we are done.

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shalomrav

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shalomrav

#14 h May 16, 2020, 6:37 PM • 2 Y

Y by Mango247, Mango247

TheUltimate123 wrote:

Let $P(x,y)$ denote the
If there is no $z\ne2019$ with $f(z)=2019$ , then $x+f(x)=2019$ for all $x$ , so $f(x)=2019-x$ .

Isn't it $f(z)=0$ ?

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fukano_2

492 posts

fukano_2

#15 h Nov 28, 2020, 1:42 AM

Y by

Solved with nprime06

We claim the solutions are $f\equiv c$ , $f(x)\equiv 2019-x$ , and $f(x)=c \text{ if }x= 0 \text{, else }f(x)=0,$ and it's easy to check that they work.

First by $P(2019,y)$ we see that $f(2019+yf(2019))=f(2019)$ . If $f(2019)\ne 0$ , then $2019+yf(2019)$ is surjective and hence $f$ is constant. Henceforth assume $f(2019)=0$ and that $f$ is not constant.

Claim: $f$ is injective at $2019$ .
Proof. FTSOC suppose $\exists\text{ } a\ne 2019\in\mathbb R$ st. $f(a)=0$ . $P(a,y)\implies f(a+yf(a))+f(ay)=f(a)+f(2019y)\iff f(ay)=f(2019y)\quad (\spadesuit).$ Then, $\begin{align*}f(ax+yf(ax))\stackrel{P(ax,y)}=&f(ax)+f(2019y)-f(axy)\stackrel\spadesuit=f(2019x)+f(2019y)-f(2019xy)\\\stackrel{P(2019x,y)}=&f(2019x+yf(2019x))\quad (\clubsuit). \end{align*}$ Since $f$ is non constant, $\exists \text{ }b\ne 1$ st. $f(ab)=f(2019b)\ne 0$ . Then by replacing $x$ with $b$ in $\clubsuit$ and varying $y$ we see that $yf(ab)$ is surjective; hence $f$ is periodic with period $(2019-a)b$ . Let $0\ne c=(2019-a)b\implies f(x+c)=f(x)\quad (\heartsuit)$ . Next, $\begin{align*}f((x+c)y)\stackrel{P(x+c,y)}=&f(x+c)+f(2019y)-f(x+c+yf(x+c))\\ \stackrel\heartsuit=&f(x)+f(2019y)-f(x+yf(x))\stackrel{P(x,y)}=f(xy)\iff f(z+cy)=f(z). \end{align*}$ However, varying $y$ over the reals we see that $f$ is constant, our desired contradiction. Hence, $a=2019,b=1,$ and $c=0$ which gives us the solution $f=\begin{cases} c\ne 0 & x=0 \\ 0 & x\ne 0 \end{cases}$ . $\square$

By the Claim, we have $P(x,1)\implies f(x+f(x))=f(2019)\implies x+f(x)=2019\implies f(x)=2019-x$ is our last solution to the FE. Having found all the solutions, we are done. $\blacksquare$

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amuthup

779 posts

amuthup

#16 h Dec 24, 2020, 4:32 AM • 2 Y

Y by Aryan-23, tiendung2006

Does this work? This uses the same idea as APMO 2019/5.

The only solutions are $f\equiv 2019-x,$ $f\equiv c,$ and $f(x)\equiv\begin{cases}0&x\ne 0\\ c&x=0\end{cases}.$ It is easy to check that these work, so assume $f$ is not one of these.

Let $P(x,y)$ denote the assertion.

$\textbf{Claim: }$ $f(2019)=0$

$P(2019,y)$ gives $f(2019+yf(2019))=f(2019).$ If $f(2019)\ne 0,$ then $2019+yf(2019)$ takes all real values as $y$ varies, so $f$ is constant. This contradicts our original assumption, so $f(2019)=0.$ $\blacksquare$

Now let $S$ be the set of reals $c$ satisfying $f(cx)=f(x)$ for all real $x.$

$\textbf{Claim: }$ $S$ is nonempty

$\emph{Proof: }$ $P(x,0)$ gives $f(x+f(x))=0$ for all $x.$ Therefore, if the only root of $f$ is $2019,$ then $f\equiv 2019-x,$ contradiction. Hence there exists $a\ne 2019$ such that $f(a)=0.$

From $P(a,y)$ and $P(2019,y),$ we find that $f(ay)=f(2019y)$ for all $y.$ Thus, $f\left(2019y\cdot\frac{a}{2019}\right)=f(ay)=f(2019y).$ Since $2019y$ can take any real value, we are done. $\blacksquare$

Fix $c\in S.$ Since $c\in S\iff\frac{1}{c}\in S,$ we may assume $|c|>1.$

The assertions $P(cx,y)$ and $P(x,y)$ yield
$\begin{align*} f(cx+yf(x)) &= f(cx)+f(2019y)-f(cxy)\\ &= f(x)+f(2019y)-f(xy)\\ &= f(x+yf(x)). \end{align*}$ Therefore, $M=\frac{cx+yf(x)}{x+yf(x)}\in S.$

Choose $x\ne 0$ such that $f(x)\ne 0,$ and let $y=\frac{1}{f(x)}.$ Then, $M=\frac{cx+1}{x+1},$ so as $x$ varies, $M$ takes all values in the interval $(1,c).$ This implies that $f(a)=f(b)$ for all $b\in (a,ca),$ so $f$ must be constant, contradiction.

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Supercali

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Supercali

#17 h Dec 24, 2020, 8:14 AM • 1 Y

Y by Mango247

Really nice problem! I have a weak spot for entirely or nearly-entirely wrapped FEs on $\mathbb{R}$

For storage:

Clearly $\boxed{f(x)=c \ \ \forall x \in \mathbb{R}}$ is a solution for all values of $c$ . Henceforth assume that $f$ is non-constant.

Let $P(x,y)$ denote the original assumption. If $f(2019) \neq 0$ , then $P \left (2019,\dfrac{y-2019}{f(2019)} \right )$ implies $f(2019)=f(y)$ for all $y$ , which contradicts that $f$ is non-constant. Hence $f(2019)=0$ , and so $P(x,1)$ $\implies$ $f(x+f(x))=0$ for all $x$ . If $x+f(x)=2019$ for all $x$ , we get $\boxed{f(x)=2019-x \ \ \forall x \in \mathbb{R}}$ which is indeed a solution.
Else suppose $x+f(x)=u \neq 2019$ for some $x$ . Then $P \left (u, \dfrac{y}{2019} \right )$ $\implies$ $f(y)=f(ky)$ for all $y$ , where $k =\dfrac{u}{2019} \neq 1$ .

If $f(x)=0$ for all $x \neq 0$ , then we get $\boxed{f(x)=0 \ \text{for} \ x \neq 0 \ \text{and} \ f(0)=c}$ which is a solution for all values of $c$ . Else $f(t) \neq 0$ for some $t \neq 0$ . Comparing $P \left ( t, \dfrac{y}{f(t)} \right )$ and $P \left ( kt, \dfrac{y}{f(t)} \right )$ we get $f(kt+y)=f(t+y)$ for all $y$ $\implies$ $f$ is periodic with period $p=(k-1)t \neq 0$ .

Comparing $P \left ( 0, \dfrac{y}{p} \right )$ and $P \left ( p, \dfrac{y}{p} \right )$ we get $f(y)=f(0)$ for all $y$ , which contradicts that $f$ is non-constant.
$\blacksquare$

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CANBANKAN

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CANBANKAN

#19 h Feb 8, 2021, 6:42 PM • 3 Y

Y by Mango247, Mango247, Mango247

How is this a killer problem?

The only $f$ 's are constant $f$ , $f(x)=2019-x,$ and $f(x)=0\forall x\ne 0$ They clearly work.

Let $P(x,y)$ denote the assertion $f(x + yf(x)) + f(xy) = f(x) + f(2019y).$

We prove a lemma:

Lemma: If $f(y)=f(Cy)\forall y$ for some constant $C\ne 1,$ then $f(x)=0\forall x\ne 0$ , or $f$ is constant.

Proof. Suppose $f(x)\ne 0,$ for some $x\ne 0$

$P(x,y)$ gives $f(x+yf(x))=f(x)+f(2019y)-f(xy)$

$P(x,Cy), P(Cx,y)$ gives $f(x+Cyf(x))=f(x)+f(2019Cy)-f(Cxy)=f(x)+f(2019y)-f(xy)=f(x+yf(x)).$ Similarly $f(Cx+yf(x))=f(x+yf(x))$

Let $a=yf(x).$ Then we can see $f(x+a)=f(x+Ca),$ and the range of $a$ is $\mathbb{R}.$ Hence, $f(x+a)=f(\frac xC+a)$ for all $a.$ This implies $f$ is periodic with period $|x-\frac xC|=T$

Now, $P(z+T,y)-P(z,y)$ suggest $f((z+T)y)=f(zy)$ for all $y,$ which implies $f$ is constant because there are no restrictions on $z,y$ other than them being reals.

If $f(y)=0\forall y\ne 0,$ then we can see $f(0)$ can be anything.

Suppose $f(a)=0.$ Then $P(a,y)$ yields $f(a+0)+f(ay)=f(a)+f(2019y),$ or $f(ay)=f(2019y).$ If $a\ne 2019,$ we are done by our lemma, so $f(2019)=0.$

Also, $P(x,1)$ gives $f(x+f(x))=0.$ If $f(a)=0,$ we can see $a=2019$ is forced, so $x+f(x)=2019,$ or $f(x)\equiv 2019-x.$

This post has been edited 3 times. Last edited by CANBANKAN, Feb 9, 2021, 4:42 AM

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Eyed

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Eyed

#20 h Mar 12, 2021, 4:13 PM

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The only solutions are $f(x) = c$ for a constant $c$ and $f(x) = 2019 - x$ . These both obviously work.

Let $P(x,y)$ denote the assertion. Plug in $P(2019, y)$ to get $f(2019 + yf(2019)) = f(2019)$ . If $f(2019)\neq 0$ , then $2019+yf(2019)$ ranges over all real numbers, so $f(x)$ is constant. Therefore, assume $f(2019) = 0$ . Then, $P(x,1)$ gives $f(x+f(x)) = f(2019) = 0$ .

Consider if there existed some $a\neq 2019$ such that $f(a) = 0$ , and $f$ was not a constant function. By $P(a,y)$ , it gives
$f(a+yf(a)) + f(ay) = f(a) + f(2019y) \Rightarrow f(ay) = f(2019y)$ This means $f(x) = f(\frac{a}{2019}x)$ . Now, by $P(\frac{a}{2019}x, y)$ , it gives
$f\left(\frac{a}{2019}x + yf\left(\frac{a}{2019}x\right)\right) + f\left(\frac{a}{2019}xy\right) = f\left(\frac{a}{2019}x\right) + f(2019y)$ $\Rightarrow f\left(\frac{a}{2019}x + yf(x)\right) + f(xy) = f(x) + f(2019y)$ By $P(x,y)$ , this means $f(\frac{a}{2019}x + yf(x)) = f(x + yf(x))$ . Consider some fixed $x$ such that $f(x) \neq 0$ (exists by non-constant function). This means that, since we can vary $y$ , we must have $f(r) = f\left(r + \frac{2019-a}{2019}x\right)$ , so $f$ is a periodic function. Let the period be $k$ , so $f(r) = f(r+k)$ .

By $P(k, y)$ , it gives
$f(k+yf(k)) + f(ky) = f(k) + f(2019y) \Rightarrow f(yf(0)) + f(ky) = f(0) + f(2019y)$ By $P(0, y)$ , we get $f(yf(0)) = f(2019y)$ . Substituting into the above equation,w e get $f(ky) = f(0)$ , so $f$ is a constant function, a contradiction.

Therefore, assume the only value of $a$ such that $f(a) = 0$ is $a = 2019$ . Since $f(x+f(x)) = 0$ , this means $x + f(x) = 2019$ , so $f(x) = 2019-x$ . These are our only solutions.

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rama1728

800 posts

rama1728

#21 h Oct 3, 2021, 6:18 AM

Y by

CantonMathGuy wrote:

Determine all functions $f: \mathbb{R} \to \mathbb{R}$ satisfying
$f(x + yf(x)) + f(xy) = f(x) + f(2019y),$ for all real numbers $x$ and $y$ .

To begin with, we would try to extract the constant solutions. Clearly, any constant $f\equiv c$ works, now assume that $f$ is non-constant. We shall now prove a claim.

Claim. $f(z)=0\iff z=2019$ or $f(x)= 0\iff x\neq 0$ .
Proof. Note that $P(z,y)$ gives us that $f(z+yf(z))+f(zy)=f(z)+f(2019y)$ implying that $f(zy)=f(2019y)$ . This implies that $f(kx)=f(x)$ for all $x\in\mathbb{R}$ where $k=\frac{z}{2019}$ . Then, $P(kx,y)$ gives us that $f(kx+yf(x))+f(xy)=f(x)+f(2019y)$ This implies that $f(kx+yf(x))=f(x+yf(x))$ implying that $f$ is periodic with period $T=(k-1)x$ . Now, let $a$ be such that $f(a)\neq0$ and put $T=(k-1)a$ which is fixed. Then, $P(x+T,y)$ and $P(x,y)$ together give us that $f(xy+yT)=f(xy)$ and this holds for every $x$ and $y$ . This forces $T=0$ otherwise $f$ is constant. Therefore, $k=1$ , i.e. $z=2019$ or $a=0$ , i.e. $f(x)=0$ for all $x\neq0$ , because if $z\neq2019$ , then no other $a$ shall work, implying that the only such $a$ is $0$ .

Therefore, we have another "almost" constant solution, that is $f(x)=0$ for all $x\neq0$ . Now assume $f(2019)=0$ . Then, $P(x,1)$ yields $f(x+f(x))=0$ implying that $f(x)+x=2019$ or $f(x)=2019-x$ . Therefore, the only solutions are $f\equiv c$ for any real constant $c$ , $f(x)=0$ for all $x\neq0$ and $f(x)=2019-x$ for all $x\in\mathbb{R}$ .

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ZETA_in_olympiad

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ZETA_in_olympiad

#22 h Jun 3, 2022, 6:52 AM

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Let $P(x,y)$ be the given assertion. Clearly any constant function works so assume otherwise. Note that $f(2019)=0$ from $P(2019,x).$

$P(x,1)$ yields $f(x+f(x))=0$ which means $f(x)=2019-x$ if injectivity at zero, this works. Assume $f(c)=0$ for some $c\neq 2019.$ $P(c,x/2019)$ gives $f(x)=f(cx/2019).$ Moreover $P(0,x)$ implies $f(xf(0))=f(2019x).$

We may have $f(x)=0$ for all non zero $x$ and $f(0)=k$ , this works. If not then, comparing $P(x,y)$ with $P(cx/2019,y)$ gives $f(cx/2019+yf(x))=f(x+yf(x)).$ This implies $f$ is periodic with some period $p.$ But then $P(p,y)$ gives $f$ is constant, contradiction.

This post has been edited 1 time. Last edited by ZETA_in_olympiad, Jun 3, 2022, 6:53 AM

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AwesomeYRY

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AwesomeYRY

#23 h Sep 17, 2022, 2:24 AM

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We claim the solutions are $f\equiv c$ where $c$ is some constant or $f\equiv -x+2019$ . They clearly work.

Let $P(x,y)$ be the assertion. Consider $P(2019,y)$ :
\begin{equation}
f(2019+ y \cdot f(2019) = f(2019)
\end{equation}
Either $f(2019)\neq 0$ , in which case $f$ is constant, or $f(2019)=0$ . Henceforth, we assume $f(2019)=0$ and that $f$ is not constant.

Next, note that if $f(z)=0$ , then $P(z,y)$ tells us that $f(zy) = f(2019y)$ . Thus, if we let $r=\frac{z}{2019}$ , then $f(rx) = f(x)$ for all $x$ . Note that multiplying $x$ by $r$ only changes the $f(x+yf(x))$ term in $P(x,y)$ . Thus, since $f$ is non-constant, we may select $\alpha$ such that $f(\alpha)=C\neq 0$ . Thus, we have $f(\alpha r + y\cdot C) = f(r+ y\cdot C)$ , and $f$ is periodic with period $T = \alpha r - r$ .

This finally derives a contradiction since if we increase $x$ by $T$ , the only value that changes in $P(x,y)$ is $f(xy)$ . Then, by selecting $y = \frac{\delta }{T}$ , $f$ is periodic with all periods, so $f$ must be constant, contradiction.

Finally, consider if the only such $z$ is 2019. Then, note that $P(x,1)$ gives $f(x+f(x)) = 0$ , so this implies that $x+f(x) = 2019$ for all $x$ , recovering $f(x)\equiv -x+2019$ , and we have shown that the stated functions are the only possible ones. $\blacksquare$ .

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IAmTheHazard

5000 posts

IAmTheHazard

#24 h Jan 20, 2023, 9:51 PM

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The answer is $f \equiv c$ , $f(x)=2019-x$ , and $f(x)=0$ if $x \neq 0$ with $f(0)\neq 0$ arbitrary. The first two clearly work. For the third solution, note that if $x$ is nonzero, then the LHS is $f(xy)$ and the RHS is $f(2019y)$ , which are equal. If $x$ is zero, then the LHS is $f(yf(0))+f(0)$ and the RHS is $f(0)+f(2019y)$ . Since $f(0) \neq 0$ these two sides are also equal. We now prove that these are the only solutions; let $P(x,y)$ denote the given assertion.

Clearly every constant solution works, so assume that $f$ is nonconstant. From $P(2019,y)$ , we find that $f(2019+yf(2019))=f(2019)$ for all $y \in \mathbb{R}$ . If $f(2019)\neq 0$ , then this means that $f(x)=f(2019)$ for all $x$ , so $f$ is constant—contradiction. Thus $f(2019)=0$ .
Suppose that $f(a)=0 \implies a=2019$ , i.e. $f$ has a single root. From $P(x,1)$ we obtain $f(x+f(x))=0$ for all $x$ , hence $x+f(x)=2019 \implies f(x)=2019-x$ for all $x$ , which is one of our solutions.
Thus suppose that $f$ has some other root $r$ . From $P(r,x)$ we obtain $f(rx)=f(2019x)$ , so $f(x)=f(kx)$ where $k=\tfrac{r}{2019}$ . If there does not exist $x \neq 0$ with $f(x) \neq 0$ , then we get the third solution as described. Thus suppose that such an $x$ exists. From $P(x,y)$ and $P(kx,y)$ we find that
$f(kx+yf(x))=f(kx+yf(kx))=f(kx)+f(2019y)-f(kxy)=f(x)+f(2019y)-f(xy)=f(x+yf(x)).$ Since $f(x) \neq 0$ , $yf(x)$ spans the real numbers. Furthermore, $kx \neq x$ as $x \neq 0$ , so $f$ is in fact periodic with period $(k-1)x:=T \neq 0$ .
Now consider some $x \in \mathbb{R}$ (not related to $\tfrac{T}{k-1}$ ). From $P(T,x)$ , we obtain
$f(T+xf(T))+f(xT)=f(T)+f(2019x) \implies f(xf(0))+f(xT)=f(0)+f(2019x).$ On the other hand, $P(0,x)$ yields
$f(xf(0))+f(0)=f(0)+f(2019x),$ hence $f(xT)=f(0)$ . Since $T \neq 0$ , this means that $f$ is constant for all reals: contradiction. Thus we are done. $\blacksquare$

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kamatadu

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kamatadu

#25 h Jan 2, 2024, 7:46 PM

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We change the $2019$ to $2023$ which still works.

The solutions are $f\equiv c$ and $f \equiv 2023 - x$ .

Assume $f\not\equiv c$ . So there exists a $p$ such that $f(p) \neq 2023$ .

$P(2023,y) \implies f(2023 + yf(2023)) = f(2023)$ .

If $f(2023) \neq 0$ , then choosing $y \rightarrow\dfrac{y}{f(2023)} - 2023$ gives $f\equiv c$ , contradiction.

Otherwise $f(2023) = 0$ .

Now $P(0,y) \implies f(yf(0)) = f(2023y) \implies f(f(0)) =f(2023) = 0$ .

If $f(0) \neq 2023$ , then $f(y) = f\left(\frac{f(0)y}{2023}\right) = f(ry); \; r\neq 1$ .

Now $P(x,1) \implies f(x + f(x)) = 0$ .

Furthermore, $P(x + f(x),y) \implies f(y(x+f(x))) = f(2023y). \qquad\ (\clubsuit)$

Now in the main equation, comparing $P(x,y)$ and $P(rx,y) \implies f(rx+yf(x)) = f(x+yf(x))$ .

Thus $k = \dfrac{rx + yf(x)}{x + yf(x)} \implies y = \dfrac{x(k-1)}{f(x)(r-k)}$ for some $f(x) \not\equiv 0$ .

Then we can similarly get $f\equiv c$ , contradiction.

Thus we must have that $f$ is injective or $f(0) = 2023$ .

If $f$ is injective, then $(\clubsuit)$ gives that $f(x) = 2023 - x$ . Otherwise $f(0) = 2023$ .

Now putting $y=1$ in $(\clubsuit)$ , we get $f(x+f(x)) = f(2023) = 0$ .

Now comparing $P(x+f(x),y)$ and $P(2023,y)$ , we get $f((x+f(x))y) = f(2023y)$ and we can finish similarly by assuming $r = \dfrac{2023}{x+f(x)} \neq 1$ and then using APMO trick as earlier we are done. :yoda:

This post has been edited 2 times. Last edited by kamatadu, Jan 2, 2024, 7:47 PM

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john0512

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john0512

#26 h Feb 8, 2024, 6:34 PM

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$f(x+yf(x))+f(xy)=f(x)+f(2019y)$
The answer is $f(x)=c$ and $f(x)=2019-x$ , and $f(x)=0$ on $x\neq 0$ for any $f(0)$ , which clearly work. Plugging $x=2019$ , we have $f(2019+yf(2019))=f(2019).$ If $f(2019)\neq 0$ , then $2019+yf(2019)$ can be any real number, which implies that $f$ is constant. Therefore, if $f$ is nonconstant, then $f(2019)=0$ . Assume this from now on.

Plugging $y=1$ , we have $f(x+f(x))=0.$ If $2019$ is the only value of $x$ for which $f(x)=0$ , we would be done as we must have $x+f(x)=2019$ .

Thus, from now on assume there exists $r\neq 2019$ with $f(r)=0$ . Seeing the abundance of $f(x)$ , assert $x=r$ to get $f(ry)=f(2019y).$ Thus, $f(y)=f(\frac{r}{2019}y).$ This means that $f$ is "exponentially periodic". For periodic functions, a common strategy is to replace $x$ with its next cycle, so $x\rightarrow \frac{r}{2019}x$ (not changing $f(x)$ or $f(xy)$ ) gives $f(\frac{r}{2019}x+yf(x))+f(xy)=f(x)+f(2019y).$ Thus, $f(x+yf(x))=f(\frac{r}{2019}x+yf(x)).$ Assuming $f$ is not always zero on nonzero inputs (we already found that as a solution), plugging in any nonzero non-root of $f$ gives us that $f$ is periodic with a nonzero period (additively), since $yf(x)$ can be any real.

Doing the same periodic trick, $x\rightarrow x+P$ where $P$ is the period, $f(x+yf(x))$ does not change since $x+yf(x)$ increases by $P$ , and $f(x)$ and $f(2019y)$ also do not change. Thus, $f(xy)=f(xy+Py).$ Plugging $x=\frac{1}{y}$ , we have $f(1)=f(1+Py),$ so $f$ is constant as $Py$ can be any nonzero real and we are done.

remark: the $f(0)$ not having to equal 0 is quite tricky, and I initially fakesolved this. The error was that when claiming that $f(x+yf(x))=f(\frac{r}{2019}x+yf(x))$ implies that $f$ is periodic for $f(x)\neq 0$ , one has to make sure that the period is not zero, but when $x=0$ , the period is 0. So this only shows that when $f$ is nonzero on a nonzero input, $f$ is periodic.

This post has been edited 2 times. Last edited by john0512, Feb 8, 2024, 10:20 PM
Reason: fixed sol

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Pyramix

419 posts

Pyramix

#27 h Apr 6, 2024, 10:18 AM

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$f\left(x+yf\left(x\right)\right)+f\left(xy\right)=f\left(x\right)+f\left(cy\right)$
$f\left(x+f\left(x\right)\right)=f\left(c\right)$
$f\left(c+yf\left(c\right)\right)=f\left(c\right)$
So, if $f$ is injective $f\left(c\right)=0$ and $f\left(x\right)+x=c$ which means $f\left(x\right)=c-x$ .
So, all we need is to prove that function is injective.

Suppose $f\left(c\right)\neq0$ . Then, $f\left(c+y\right)=f\left(c\right)$ and the function is a constant. (Any constant is fine.)
So, we have $f\left(c\right)=0$ . Hence, $f\left(x+f\left(x\right)\right)=0$ . We need injectivity at 0 to finish.
Suppose $f\left(a\right)=f\left(b\right)=0$ for distinct $a,b$ . Then, plugging $P\left(a,y\right)$ we get $f\left(ay\right)=f\left(cy\right)=f\left(by\right)$ . Let $r=\frac{a}{b}$ . Let $S$ be the set of all $r$ for which $f\left(x\right)=f\left(rx\right)$ . Then, plug $P\left(x,r\right): f\left(x+rf\left(x\right)\right)=0$ .
Note, plugging $P\left(rx,y\right)$ gives $f\left(rx+yf\left(x\right)\right)=f\left(x+yf\left(x\right)\right)=f\left(x+rf\left(x\right)\right)$

This means that $f(rx+yf(x))=f(x+yf(x))$ . Take $f(x)\ne0$ , and note that $f$ is a periodic function with period $(r-1)x$ as $yf(x)$ can be arbitrary. Let $T=(r-1)x$ . Then, $f(x)=f(x+T)$ . Plug $P(x+T,y)$ to get that $f((x+T)y)=f(xy)$ . Hence, if $T$ is the period then so is $yT$ . Since $y$ can be arbitrarily small, we get that $f$ is a constant for all numbers other than 0.

If $f(x)=0$ for every $x\ne0$ then we can have $f(0)$ and still original FE works.

So, solutions are:

f(x)=c-x;
f(x) = constant
f(x)=0 for $x\ne0$ and $f(0)$ is arbitrary.

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GrantStar

812 posts

GrantStar

#28 h Aug 30, 2024, 9:58 PM • 2 Y

Y by GeoKing, OronSH

Answers: $f \equiv c$ , $f(x)=2019-x$ , and $f(0)=c$ with $f(k)=0$ for $k\neq 0$ .

Let $P(x,y)$ denote the assertion. First, $x=2019$ gives $f(2019+yf(2019))=f(2019)$ so $f$ is constant or $f(2019)=0.$ If $f(0)=0$ , then $P(0,y)$ gives $f \equiv 0$ , so assume that $f(0)\neq 0$ . Now, $y=1$ gives $f(x+f(x))=0$ .

If no nonzero real number is sent to a nonzero number, we obtain the third solution set. If $2019$ is the only number sent to $0$ then $f(x+f(x))=0$ gives our $f(x)=2019-x$ solution
Thus suppose $f(a)=0$ for $a \neq 0, 2019$ ; we will show no more solutions exist. $P(a,y)$ now gives that $f(ay)=f(2019y)$ , or $f(x)=f(\frac{2019}{a}x)$ .

Claim: Here, $f$ is periodic with period $d$ .
Proof. There exists a nonzero $d$ with $f\left (\frac{d}{\frac{2019}{a}-1}\right)\neq 0$ . Then, let $m-n=d$ . $P(\frac{2019}{a}x,y)$ gives $f(\frac{2019}{a}x+yf(x))=f(x+yf(x))$ . Letting $x=d$ there exists a $y$ with the LHS as $m$ and RHS as $n$ as desired. $\blacksquare$

Now, let $a\neq b$ . Let $y, x_1, x_2$ be reals with $d \mid x_1-x_2$ and $a=yx_1$ , $b=yx_2$ . $P(x_1,y)$ and $P(x_2,y)$ give that $f(a)=f(b)$ since all terms except the $f(xy)$ are the same.

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joshualiu315

2513 posts

joshualiu315

#29 h Aug 31, 2024, 5:19 AM

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The answer is

$f(x) = \boxed{\begin{cases} 2019-x \\ c, \ c \in \mathbb{R} \\ 0, \ x \neq 0 \ \text{ and } \ f(0) = c \end{cases}}$
which all work. Denote the given assertion as $P(x,y)$ .

Notice that $f$ can be constant, so henceforth assume otherwise. Observe that $P(2019,y)$ gives

$f(2019+yf(2019)) = f(2019).$
If $f(2019) = k \neq 0$ , then the RHS is always equal to $k$ , making $f$ constant. Hence, $f(2019) = 0$ .

Then, $P(x,1)$ yields

$f(x+f(x))=0.$
Suppose that $x+f(x) = 2019$ . We easily see that $f(x) = 2019-x$ is a solution, so assume otherwise from now on.

Let $x+f(x) = r$ such that $r$ is another root of $f$ . Plug in $P(r,y)$ to get

$f(ry) = f(2019y) \implies f(y) = f\left( \frac{r}{2019} y\right).$
Obviously, if $f(x) = 0$ for all values except $x = 0$ , this is a solution. Else, assume there is a value $s \neq 0$ such that $f(s) \neq 0$ . Comparing $P(s,\tfrac{y}{f(s)})$ and $P(\tfrac{r}{2019}s,\tfrac{y}{f(s)})$ shows that

$f(s+y) = f\left(\frac{r}{2019} s + y \right).$
This shows the periodicity of $f$ so we let the period be $p$ . Comparing $P(0,\tfrac{y}{p})$ and $P(p,\tfrac{y}{p})$ yields $f(y) = f(0)$ for all $y$ , contradicting the nonconstant condition.

We have exhausted all of the cases, so we are done.

Remark: We apply the $x \neq 0$ condition while proving periodicity, because otherwise the period would equal $0$ .

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pie854

243 posts

pie854

#30 h Sep 4, 2024, 11:09 PM

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Clearly $f(x)\equiv k$ for some constant $k$ satisfies our equation, so let's assume that $f$ is non-constant. Let $P(x,y)$ denote the given equation.

$P(0,x)$ implies that $f(xf(0))=f(2019x)$ . Next we have, $P(f(0),x)\implies f(f(0)+xf(f(0))) = f(f(0))$ this means that $f(f(0))=0$ since otherwise $f$ is constant. It also means $f(2019)=0$ . So $P(x,1)$ gives us $f(x+f(x))=0$ and so $P(x+f(x),y)\implies f((x+f(x))y)=f(2019y) \qquad (1)$ Now we will prove a result which will help us "unwrap" $f$ and also finish the problem.

Claim: If $f(tx)=f(x)$ for all $x$ , then $t=1$ .
Proof. Then, we can check that $P(t,y)$ and $P(1,y)$ gives us $f(t+yf(1))=f(2019y)=f(1+yf(1)).$ Note that this equation itself implies $f(1)\neq 0$ (otherwise $f(2019y)=0$ , a contradiction). Thus we get $f(x)=f(x+t-1)$ for all real $x$ . Now $P(t-1,x)\implies f(t-1+xf(0))+f(x(t-1))=f(0)+f(2019x) \implies f(xf(0))+f(x(t-1))=f(0)+f(2019x)$ while $P(0,x)$ implies $f(xf(0))+f(0)=f(0)+f(2019x)$ and so we get $f(x(t-1))=f(0)$ for all $x$ . This is a contradiction unless $t=1$ . So we're done. ////

Applying the claim for $(1)$ we get that $x+f(x)=2019$ for all $x$ , thus $f(x)=2019-x$ . Conversely it's easy to check this function works.

Edit: oops, seems like I missed the solution $f(x)=\begin{cases} 0 & x\neq 0 \\ k & x=0\end{cases}$

This post has been edited 1 time. Last edited by pie854, Sep 4, 2024, 11:17 PM

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SomeonesPenguin

123 posts

SomeonesPenguin

#31 h Sep 17, 2024, 10:02 AM

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The solutions are constant $f$ , $f(x)=2019-x$ and $f(x)\equiv\begin{cases}0&x\ne 0\\ c&x=0\end{cases}$ these clearly work. We will suppose that $f$ isn't of the first and third from, hence there exists $x\neq 0$ such that $f(x)\neq 0$ .

Let $P(x,y)$ denote the given assertion.

$\bullet$ $P(2019,x)\Rightarrow f(2019+yf(2019))=f(2019)$ , hence $f(2019)=0$ since $f$ isn't constant.

$\bullet$ $P(x,1)\Rightarrow f(x+f(x))=0$

Now suppose that there exists $a\neq 2019$ such that $f(a)=0$ . $P(a,y)$ gives $f(ay)=f(2019y)$ . Now subtracting $P(ax,y)$ from $P(2019x,y)$ gives $f(ax+yf(2019x))=f(2019x+yf(2019x))$
Therefore, fixing $x\neq 0$ such that $f(2019x)\neq 0$ (notice that $x\neq 0$ from our assumption at the beginning) and $y\mapsto \frac{y}{f(2019x)}$ gives $f(y+2019x)=f(y+ax)$ , $\forall y\in \mathbb R$ and since $a\neq 2019$ we get $f(x)=f(x+c)$ , $\forall x\in\mathbb R$ , where $c=|x(2019-a)| \neq 0$ . Now subtracting $P(x+c,y)$ from $P(x,y)$ gives $f(xy)=f(xy+yc)$ so $x\mapsto \frac{a}{y}$ gives $f(a+yc)=0$ , $\forall y\in\mathbb R\setminus \{0\}$ so $f\equiv 0$ .

Finally, $f$ is injective at $0$ so from $f(x+f(x))=0=f(2019)$ we get $f(x)=2019-x$ . $\blacksquare$

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N3bula

254 posts

N3bula

#32 h Oct 12, 2024, 12:37 AM

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Let $P(x, y)$ denote assertion.
$P(2019, y)$
$f(2019+yf(2019))=f(2019)$ Thus we get either $f(2019)=0$ or $f(x)=c$ for some fixed $c$ . Suppose $f(2019)=0$ .
$P(x, 1)$
$f(x+f(x))=0$ Suppose $f(k)=0$ and $k\neq 2019$ .
$P(k, y)$
$f(ky)=f(2019y)$ Thus if we let $m=\frac{k}{2019}$ we get $f(x)=f(mx)$ .
$P(mx, y)$
$f(mx+yf(x))=f(x)+f(2019y)-f(xy)=f(x+yf(x))$ Thus this implies that $f$ is periodic, now let the period be $t$ .
$P(x+t, y)-P(x, y)$
$f((x+t)y)=f(y)$ Which implies that for all $x\neq 0$ $f(x)=0$ . Thus we get either $f$ is constant, $f$ is $0$ for everything that is not
$0$ or $f(x)=2019-x$ .

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bin_sherlo

662 posts

bin_sherlo

#34 h Oct 12, 2024, 10:39 AM

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$f(x + yf(x)) + f(xy) = f(x) + f(2019y)$ Answers are $f\equiv c$ and $f(x)=2019-x$ and $f(x)=0$ for all $x\neq 0$ and $f(0)$ is any real number. Constant functions clearly satisfy the equation hence assume that $f$ is nonconstant.

Case $I$ : $f$ is periodic.
Let $T$ be the period. Comparing $P(x,y)$ with $P(x+T,y)$ implies $f(xy)=f(xy+Ty)$ .Plug $\frac{x}{y},\frac{y}{T}$ to get $f(x)=f(x+y)$ for $y\neq 0$ which is obvious for $y=0$ . So $f$ is constant which contradicts with our initial assumption. $\square$

Case $II$ : $f$ is not periodic.

Claim: $f(2019)=0$ .
Proof: If $f(2019)\neq 0,$ then plugging $P(2019,\frac{y-2019}{f(2019)})$ gives $f(y)=f(2019)$ but this is impossible since $f$ is non-constant. $\square$

Claim: For $x\neq 0,$ we have $f(x)\in \{0,2019,2019-x\}$ .
Proof: Suppose that $f(x)\neq 0,2019$ .
$P(x,\frac{x}{2019-f(x)}): \ f(x+\frac{xf(x)}{2019-f(x)})+f(\frac{x^2}{2019-f(x)})=f(x)+f(\frac{2019x}{2019-f(x)})$ Since $x+\frac{xf(x)}{2019-f(x)}=f(\frac{2019x}{2019-f(x)}),$ we conclude that
$f(x)=f(\frac{x^2}{2019-f(x)})$ We compare $P(x,y)$ with $P(\frac{x^2}{2019-f(x)},y)$ in order to verify
$f(x+yf(x))=f(\frac{x^2}{2019-f(x)}+yf(x))$ Now plug $x,\frac{y-x}{f(x)}$ which gives us the equation
$f(y)=f(y+\frac{x^2}{2019-f(x)}-x) \ \ \text{for} \ f(x)\neq 0,2019$ Since $f$ is not periodic, we must have $\frac{x^2}{2019-f(x)}=x$ . We have assumed that $x\neq 0$ so $x=2019-f(x)\iff f(x)=2019-x$ which proves our claim. $\square$

If there exists $f(a)=0$ where $a\neq 2019,$ then choose $P(a,\frac{y}{2019})$ which yields $f(y.(\frac{a}{2019}))=f(y)$ . Since $f$ is nonconstant, we can choose an $x$ such that $f(x)\neq 0$ . Comparing $P(x,y)$ with $P(\frac{xa}{2019},y)$ gives us
$f(x+yf(x))=f(\frac{xa}{2019}+yf(x))$ By plugging $y\rightarrow \frac{y-x}{f(x)},$
$f(y)=f(y+\frac{xa}{2019}-x)$ Since $f$ is not periodic, we get that $xa=2019x\iff x(a-2019)=0$ . So $x=0$ and this means if $x\neq 0,$ then $f(x)=0$ which is a solution.

If $f(b)=2019$ and $b\neq 0,$ then plug $P(x,1)$ to get $f(x+f(x))=0$ . Choosing $x=b$ gives $f(b+2019)=0$ . But by looking at the previous case, if $f(b+2019)=0$ and $b\neq 0,$ then we get the function $f(x)=0$ for all $x\neq 0$ .

Now, we have worked on the cases $f(a)\in \{0,2019\}$ for $a\neq 2019,0$ respectively. We have $f(x)=2019-x$ for $x\neq 0$ .
$P(0,1): \ f(yf(0))=f(2019y)=2019-2019y$ Note that $f(0)\neq 0$ . So put $y=1$ to verify $2019-f(0)=f(f(0))=0$ thus, $f(0)=2019$ . We get the function $f(x)=2019-x$ for all reals $x$ as desired. $\blacksquare$

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OronSH

1720 posts

OronSH

#35 h Nov 12, 2024, 4:20 AM • 1 Y

Y by megarnie

Answers are $f(x)=c,f(x)=2019-x,f(x)=0\forall x\ne 0$ , which we can check work.

First $P(2019,y)$ gives $f(2019+yf(2019))=f(2019)$ , so either $f$ constant or $f(2019)=0$ .

If $f$ is injective at $0$ then $P(x,1)$ gives $f(x+f(x))=0$ implies $f(x)=2019-x$ .

Otherwise suppose $f(2019s)=0$ for $s\ne 1$ . Now $P(2019s,\tfrac x{2019})$ gives $f(x)=f(sx)$ for all $x$ .

Next $P(sx,y)$ gives $f(sx+yf(x))=f(sx+yf(sx))=f(sx)+f(2019y)-f(sxy)=f(x)+f(2019y)-f(xy)=f(x+yf(x)).$ If $f(x)\ne 0$ , then $f$ has period $p=(s-1)x$ .

If $x\ne 0$ then $p\ne 0$ and $P(x+p,y)$ gives $f(xy+py)=f(x+p)+f(2019y)-f(x+p+yf(x+p))=f(x)+f(2019y)-f(x+yf(x))=f(xy).$ This implies $f$ constant.

Otherwise $f(x)\ne 0\implies x=0$ . This gives the answers.

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megarnie

5532 posts

megarnie

#36 h Dec 1, 2024, 2:35 PM • 1 Y

Y by OronSH

Did this a while back but never posted solution

The only solutions are $f \equiv c$ for some constant $c$ , $f(x) = 2019 - x \forall x$ , and $f(x) =0 \forall x\ne 0$ where $f(0)$ is any real not equal to $0$ .

Let $P(x,y)$ be the given assertion. Clearly all constants work, so assume $f$ isn't constant.

$P(2019. y): f(2019 + y f(2019)) = f(2019)$ . If $f(2019) \ne 0$ , then $f$ is constant so $f(2019) = 0$ .

Note that $f(x) = 0 \forall x\ne 0$ always works, so assume that there is some nonzero real number that isn't a zero of $f$ .

Claim: $f$ is injective at $0$ .
Proof: Suppose not and $f(c) = 0$ for some $c\ne 2019$ .

Firstly, if $c = 0$ , then $P(0,y)$ gives $f(2019y) = 0$ , so $f$ is constant, absurd. Therefore $c\ne 0$ .

$P(c, x): f(cx) = f(2019x)$ . Let $r = \frac{2019}{c}\ne 1$ . We have $f(x) = f(rx)$ for all reals $x$ .

$P(rx, y): f(rx + y f(x)) + f(rxy) = f(rx) + f(2019y)$ . Comparing with the original FE gives $f(x + y f(x)) = f(rx + y f(x))$
Now let $t\ne 0$ satisfy $f(t) \ne 0$ . Note that $y f(t)$ is surjective over reals as $y$ varies, so $f(t + y) = f(rt + y) \implies f(x) = f(x + t(r-1)) \forall x\in \mathbb R$ Now let $d = t(r-1) \ne 0$ .

$P(x+d, y) - P(x,y): f((x+d) y) = f(xy) \implies f(xy) = f(xy + dy)$ . Setting $x =0$ gives $f(dy) = f(0)$ , so $f$ is constant, a contradiction. $\square$

$P(x,1): f(x + f(x)) = 0\implies x + f(x) = 2019 \implies f(x) = 2019 - x$ .

This post has been edited 1 time. Last edited by megarnie, Dec 1, 2024, 2:39 PM

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awesomeming327.

1665 posts

awesomeming327.

#37 h Dec 22, 2024, 7:14 PM

Y by

The answers are

$f(x)=c$ for some $c\in \mathbb R$
$f(x)=2019-x$
$f(x)=0$ for all nonzero values $x$ and $f(0)=c$ for some $c\in\mathbb R$

all of which verifiably work. From now on, assume $f$ is neither the first nor the third, and we'll prove that $f$ must be identically $2019-x$ . Let $P(x,y)$ denote the assertion $f(x + yf(x)) + f(xy) = f(x) + f(2019y)$ .

Claim: $f(2019)=0$ .

Proof. If we have $f(2019)\neq0$ then $P(2019, \tfrac{z-2019}{f(2019)})$ gives $f(z)=f(2019)$ for all reals $z$ , which is a contradiction to $f$ being nonconstant.

Claim: $f(x)=0\implies x=2019$

Proof. Suppose there exists $r\neq 2019$ such that $f(r)=0$ . Then taking $P(r,x)$ over $x\in \mathbb R\setminus 0$ gives $f(rx)=f(2019x)=c$ such that $c\neq 0$ . If such a value $x$ does not exist then $f$ is constantly zero over all values except at zero, contradiction. We have
$\begin{align*} P(rx,y)&\implies f(rx+cy)+f(rxy)=c+f(2019y) \\ P(2019x,y)&\implies f(2019x+cy)+f(2019xy)=c+f(2019y) \\ \end{align*}$ Note that $f(rxy)=f(2019xy)$ so $f(rx+cy)=f(2019x+cy)$ . Fixing $x$ and varying $y$ tells us that $f$ is periodic with some period $k$ . Then comparing $P(0,y)$ with $P(k,y)$ gives $f(yk)=f(0)$ for all $y$ . Note that from this we can easily deduce that $f$ is constant, contradiction.

Now, we have $P(x,1)$ giving $f(x+f(x))=f(2019)=0$ which implies $x+f(x)=2019$ for all $x$ giving $f(x)=2019-x$ as desired.

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HoRI_DA_GRe8

584 posts

HoRI_DA_GRe8

#38 h Jan 1, 2025, 2:04 PM • 1 Y

Y by L13832

New year solve !

RMM 2019 P5 wrote:

Determine all functions $f: \mathbb{R} \to \mathbb{R}$ satisfying
$f(x + yf(x)) + f(xy) = f(x) + f(2019y),$ for all real numbers $x$ and $y$ .

The solutions to this functional equation is,
$f(x) = \boxed{\begin{cases} 2019-x \\ c, \ c \in \mathbb{R} \\ 0, \ x \neq 0 \ , \ f(0) = c \end{cases}}$ It can be checked that these work.

First we plug in $y=1$ to get $f(x+f(x))=f(2019)$ and then, $x=2019,y=\frac{x-2019}{f(2019)}$ to get $f(x)=f(2019)=c \forall x \in \mathbb{R}$ in accordance with the condition that $f(2019) \not = 0$ .

Now consider the case when $f(2019)=0$ .Here we give a claim which forms the crux of the problem.

Claim : $f(x)$ is injective at $2019$ or $f$ is constant at all points with $f(2019=0$ .
Proof : Let $k$ be a real such that $f(k)=0$ where $k\ge 0$ and $k \not = 2019)$ .The assertion $P(k,y)$ gives that
$f(ky)=f(2019y)$ Making plausible substitution gives ,
$f(x)=f(rx)=f(r^2x)=\cdots=f(r^nx), r \in \mathbb{R}^+,r \ge 1$ Now the assertion $P(rx,y)$ gives,
$f(rx+yf(rx))+f(rxy)=f(rx)+f(2019y) \implies f(rx+yf(x))+f(xy)=f(x)+f(2019y)=f(xy)+f(x+yf(x))=f(xy)+f(rx+ryf(x))$ Plugging in $y=\frac{1}{f(x)}$ in the above equation gives,
$f(rx+1)=f(rx+r) \implies f\text{ is periodic with period r-1 }$ But for $R$ abruptly large we get $f(R+r-1)=f(R) \implies f(x)=f(\frac{R+r-1}{R}\cdot x)$ .Which gives for $s \in (1,1+e)$ for $e\ge 0$ and small, $f(sx)=f(x) \forall s$ .Which implies $f$ is constant other than at $f(0)$ which gives rise to the second solution .Otherwise any such $k$ does not exist i.e. the function is injective at $2019$ $\square$

Since the function is injective at $2019$ and we have $f(x+f(x))=f(2019)$ we get that ,
$x+f(x)=2019 \implies f(x)=2019 -x$ Thus we have obtained all the solutions for the problem and hence WE DUN $\blacksquare$

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torch

996 posts

torch

#39 h Jan 7, 2025, 4:48 AM

Y by

The only solutions are $f(x) = c$ , $f(x)=\begin{cases} 0 & x \neq 0 \\ c &x=0 \end{cases}$ , and $f(x)=2019-x$ . We have $P(2019, y) \implies f(2019+yf(2019)) = f(2019)$ . If $f(2019) \neq 0,$ then $2019+yf(2019)$ can attain any real value, implying $f$ is constant. Hence, assume $f(2019)=0$ and $f \not\equiv 0$ . Now, $P(x, 1) \implies f(x+f(x))=f(2019)=0$ . If $2019$ is the only root of $f$ , then we must must have $x+f(x)=2019 \implies f(x)=-x+2019$ . Now, assume there exists another $z \neq 0$ that is a root of $f$ . Then $P(z, y) \implies f(yz)=f(2019y) \implies f\left( \frac{yz}{2019}\right)=f(y)$ . Now, setting $P\left( \frac{xz}{2019}, y \right)$ and $P(x, y)$ equal gives us $f (\frac{xz}{2019}+yf(x))=f(x+yf(x))$ . We will now prove $f$ is periodic. Let $T=x\left(\frac{z}{2019}-1 \right)$ be the period of $f$ . Then $yf(x)$ can attain any real value, except for when $x=0$ for $f \equiv 0$ . Now, $P(x+T, y) \implies f(xy+Ty)-f(xy) \implies f(c+Ty)-f(c)$ . It follows easily that $f \equiv a$ for $x \neq 0$ . Clearly, we must have $a=0$ , yielding $f(x)=\begin{cases} 0 & x \neq 0 \\ c &x=0 \end{cases}$ as the last solution.

This post has been edited 1 time. Last edited by torch, Jan 7, 2025, 4:48 AM

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Ilikeminecraft

298 posts

Ilikeminecraft

#40 h Jan 31, 2025, 3:16 AM

Y by

so hard

The solution set is $f\equiv c$ , $f(x)\equiv 2019-x$ , and $f(x)=c \text{ if }x= 0 \text{, else }f(x)=0$ where $c \in \mathbb R.$ It is easy to check these work.

Note that $f\equiv c$ is an obvious solution. From here, assume $f$ is non-constant.

$(2019, x) \implies f(2019 + x f(2019)) + f(2019x) = f(2019) + f(2019x) \implies f(2019 + xf(2019)) = f(2019).$ Since $f$ is nonconstant, we obtain $f(2019) = 0.$

Let $z\neq 2019$ be a root of $f.$ If $z = 0,$ taking $(0, x)$ implies $f$ is a constant.

We prove that $z = 2019.$ First, take $(z, x) \implies f(kx) = f(2019x).$ Now, by subtracting $\left(\frac{2019x}{z}, \frac{c}{f(x)}\right), (x, \frac{c}{f(x)})$ to get that $f\left(\frac{2019 x}{k} + c\right) = f(x + c).$ Thus, $f$ is periodic every $p = \frac{2019}k - 1.$ Take $(x + p, y), (x, y)$ to give $f(xy + py) = f(xy).$ By keeping $xy$ constant, we can vary $y$ to get that $f$ is constant other than $0.$ This gives our 3rd solution.

If no such $z$ exists, then we get $f(x + f(x)) = 0,$ so $f(x) = 2019 - x.$

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Ilikeminecraft

298 posts

Ilikeminecraft

#41 h Jan 31, 2025, 3:17 AM

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ihatemath123

3430 posts

ihatemath123

#42 h Yesterday at 12:33 AM

Y by

For sanity reasons, let $h(x) = \tfrac{f(2019x)}{2019}$ , so that our functional equation is
$h(x+yh(x)) + h(xy) = h(x)+h(y).$ Then, the solutions are $h(x)=1-x$ , $h(x) = C$ and
$h(x) = \begin{cases}C & \text{if } x = 0 \\ 0 & \text{otherwise.}\end{cases}$ From hereon, assume $h$ is not constant.

Claim: We have $h(1) = 0$ .
Proof: Taking $P(1,x)$ gives us $h(1+xh(1)) = h(1)$ . If $h(1) \neq 0$ , we have that $h$ is constant, contradiction.

Claim: For any root $r$ , we have $h(xr) = h(x)$ for all $x$ .
Proof: This is what we get from $P(r,x)$ .

Now, we split into cases.

Suppose there exists a root $r \neq 1$ and an $x$ such that $h(x) \neq 0$ and $x \neq 0$ . Comparing $P(x,y)$ and $P(xr,y)$ gives us $h(rx+yh(x)) = h(x+yh(x)),$ so we have periodicity with period $rx-x$ . If we have a period of $p$ , comparing $P(x,y)$ and $P(x+p,y)$ for any $x$ and $y$ gives us
$h(xy) = h(xy+py),$ which implies $h$ constant, contradiction.
Suppose the only root is $r=1$ . Then, taking $P(x,1)$ gives us
$h(x+h(x)) = h(1),$ so $h(x) = 1-x$ for all $x$ and we're done.
Suppose $h(x) = 0$ for all $x=0$ . Yay, we're done!!!!!!!!!!

This post has been edited 2 times. Last edited by ihatemath123, Yesterday at 12:34 AM

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