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Source: Baltic Way 2023/4

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676 posts

#1 h Nov 11, 2023, 2:54 PM

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Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $f(f(x)+y)+xf(y)=f(xy+y)+f(x)$ for reals $x, y$ .

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#2 h Nov 11, 2023, 3:21 PM

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Quite an interesting question. Here's my partial solution:
Make x=0, y=-f(0), then we will have f(f(0)-f(0))+0f(-f(0))=f(0*-f(0)-f(0))+f(0).
Which means f(-f(0))=0
Make x=-f(0),y=0, then we will have f(0)^2=0, f(0)=0
Make y=0, then we will know f(f(x))=f(x)
(not sure about the rest)

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#3 h Nov 11, 2023, 3:48 PM • 5 Y

Y by Tintarn, Deadline, Tellocan, math_comb01, Beeoin

This problem is proposed by me. Hope you all enjoyed it!

Official Solution

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#4 h Nov 11, 2023, 9:01 PM

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I think maybe I can tidy it up a bit.

Solution

This post has been edited 1 time. Last edited by tadpoleloop, Nov 11, 2023, 9:02 PM

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eulerleonhardfan

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eulerleonhardfan

#5 h Nov 12, 2023, 7:42 AM

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Let $p(x,y)$ denote the given assertion.
Since $f(x)=0$ is a solution, suppose that $f$ is not always 0.
$p(x, \frac{f(x)}{x}):$ $f(\frac{f(x)}{x})=\frac{f(x)}{x}$ for all $x\neq 0$
$p(0, -f(0)): f(-f(0))=0$
If $f(0) \neq 0$ , then $f(\frac{f(-f(0))}{-f(0)})=\frac{f(-f(0))}{-f(0)}=0$ Let there be some real number $t \neq 0$ s.t. $f(t) \neq 0$ . Then
$f(\frac{f(\frac{f(t)}{t})}{\frac{f(t)}{t}})=f(\frac{\frac{f(t)}{t}}{\frac{f(t)}{t}})=f(1)=1$ Let there be some real number $u\neq 0$ s.t. $f(u)=0$ . This is impossible as $p(1, u): f(u+1)=u+1$ and $p(u, 1): f(u+1)=f(2u)+1$ yielding $f(2u)=u$ Taking $f$ on both sides, we get $f(u)=f(f(2u))=f(2u)=u=0$ , a contradiction.
The finishing steps are the same as the official solution.

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5583 posts

#6 h Nov 12, 2023, 2:38 PM

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Solved with vsamc

The only solutions are $\boxed{f\equiv 0}$ and $\boxed{f(x) = x}$ , which both work. Now we prove they are the only solutions. Let $P(x,y)$ denote the given assertion.

We may assume that $f$ is nonconstant since the only constant solution is clearly zero.

For $x\ne 0$ , $P\left( x, \frac{f(x)}{x} \right): x f\left( \frac{f(x)}{x} \right) = f(x)$ , so $f\left( \frac{f(x)}{x} \right) = \frac{f(x)}{x}$ . Let $f(0) = c$ .

$P(0,x): f(x + c) = f(x) + c$ , so $f(-c) = 0$ . If $c\ne 0$ , then $\frac{f(-c)}{-c} =0$ is a fixed point of $f$ , so $c = 0$ .

$P(x,0): f(f(x)) = f(x)$ .

$P(f(x), 1): f(f(x) + 1) + f(x)f(1) = f(f(x) + 1) + f(x)\implies f(x) f(1) = f(x)\implies f(1) = 1$ . Now let $f(-1) = d$ . We see that $f(d) = d$ .

$P(-1,x): f(x + d) = f(x) + d$ .

$P(x, d): f(f(x) + d) + xd = f(xd + d) + f(x)$ . Since $f(f(x) + d) = f(f(x)) + d = f(x) +d$ , we see $f(xd + d) = xd + d$ . If $d\ne 0$ , then setting $x$ to $-\frac{1}{d} - 1$ gives $f(-1) = -1$ , so $d = -1$ , which implies $f(x - 1) = x-1$ , so $f$ is the identity. Now we assume $d = 0$ .

$P(1,x): f(x +1) + f(x) = f(2x) + 1$ . Hence $f(2x) = f(x + 1) + f(x) - 1$ for each $x$ . Thus, $f(-2) = f(-1) + f(0) - 1 = -1$ . Hence $\frac{f(-2)}{-2} = 2$ , so $f(2) = 2$ .

$P(x, 2): f(f(x) + 2) + 2x = f(2x + 2) + f(x) = f(x) + f(x+1) + f(x+2) - 1$ . Plugging $x = -2$ here gives $f(-1 + 2) - 4 = f(-2) + f(-1) + f(0) - 1$ , so $-3= -2$ , absurd! Therefore, $d\ne 0$ , so $f(x) = x$ is our only nonconstant solution.

This post has been edited 2 times. Last edited by megarnie, Nov 12, 2023, 3:26 PM

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#7 h Jan 15, 2024, 10:18 AM

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We uploaded our solution https://calimath.org/pdf/BalticWay2023-4.pdf on youtube https://youtu.be/TPsMv_clprQ.

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#8 h Jun 6, 2024, 6:25 AM

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Solution

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#9 h Jul 28, 2024, 2:37 PM

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Denote $P(x,y)$ as the assertion of the following F.E.
By $P(0,-f(0))$ we get $f(-f(0))=0$ , now by $P(-f(0),x)$ we get $(1-f(0))f(0)=f(0)$ which gives $f(0)=0$ .
Now $P(x.0)$ gives $f(f(x))=f(x)$ , $P(-1,x)$ gives $f(x+f(-1))=f(x)+f(-1)$ , and for $x \ne 0$ by $P \left(x, \frac{f(x)}{x} \right)$ we get $f \left(\frac{f(x)}{x} \right)=\frac{f(x)}{x}$ . Since $f(x)=0$ is a solution suppose there exists $d$ s.t. $f(d) \ne 0$ then by $x=f(d)$ here we get $f(1)=1$ and by $P(1,x)$ we get $f(x+1)+f(x)=f(2x)+1$ , now by indooks we get $f(x+nf(-1))=f(x)+nf(-1)$ for any integer $n$ so in the previous equation set $x$ to be $x+nf(-1)$ to get that $f(x+nf(-1)+1)=f(2x)-f(x)+nf(-1)+1$ , also remember that $f(nf(-1))=nf(-1)$ and now $f(nf(-1)+1)=nf(-1)+1$ follow directly. From $P(x,nf(-1))$ we get $f(nf(-1)x)=nf(-1)x$ so if $f(-1) \ne 0$ then we get $f(x)=x$ for all reals $x$ .
So suppose otherwise that $f(-1)=0$ then $f(-2)=-1$ but this means $f(-1)=-1$ which contradicts $f(-1)=0$ .
Therefore the only solutions to this F.E. are $f(x)=x$ and $f(x)=0$ for all reals $x$ , thus done :cool:

:cool:

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Teirah

#10 h Apr 3, 2025, 5:09 AM

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nice problem : )

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#11 h Apr 3, 2025, 6:21 AM

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Let $P(x,y)$ be the assertion $f(f(x)+y)+xf(y)=f(xy+y)+f(x)$ .
$P(0,-f(0))\Rightarrow f(-f(0))=0$
$P(-f(0),0)\Rightarrow f(0)=0$
$P(x,0)\Rightarrow f(f(x))=f(x)$
$P(-1,x)\Rightarrow f(x+f(-1))=f(x)+f(-1)$
P(x,f(-1))\Rightarrow f(xf(-1))=xf(-1) $So if$ f(-1)\ne0 $we have the solution$ \boxed{f(x)=x} $, which fits. Otherwise:$ P(f(x),1)\Rightarrow f(x)f(1)=f(x)4
So if $f(1)\ne1$ we have the solution $\boxed{f(x)=0}$ , which fits. Otherwise:
$P(1,-1)\Rightarrow f(-2)=-1$
$P(-2,1)\Rightarrow-2=-1$
So no more solutions.

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