Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $$f(f(x)+y)+xf(y)=f(xy+y)+f(x)$$for reals $x, y$.
Problem
Source: Baltic Way 2023/4
Tags: algebra
SBLNuclear17
11.11.2023 18:21
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)
Kimchiks926
11.11.2023 18:48
This problem is proposed by me. Hope you all enjoyed it!
Let $P(x,y)$ denote the assertion of the given functional equation.
Claim 1: $f(0)=0.$
Proof: Note that $P(0,y)$ and $P(x,0)$ gives us the following:
\begin{align*}
f(y+f(0)) =f(y)+f(0) \\
f(f(x)) +xf(0) = f(0)+f(x)
\end{align*}Consider the first expression. Plugging $y=-f(0)$ in it yields:
$$ f(-f(0)+f(0)) = f(-f(0))+f(0) \Longrightarrow f(-f(0)) = 0 $$If we denote $-f(0)=a$, then we have $f(a)=0$. Plugging $x=a$ in the second expression gives us:
$$ f(f(a)) +af(0) =f(0)+f(a) \implies af(0)=0 $$This either means that $a=0 \implies f(0)=0$ or $f(0)=0$. In both cases, the claim is proved.
Since $f(0)=0$, the second expression becomes $f(f(x)) =f(x).$ \hfill (*)
Claim 2: $f(1)=1$ or $f(x)=0$ for all real numbers $x.$
Proof: Consider $P(x,1)$:
$$ f(f(x)+1) +xf(1) = f(x+1) + f(x) $$Replacing $x$ by $f(x)$ and using (*) leads to:
\begin{align*}
f(f(f(x))+1) +f(x)f(1) = f(f(x)+1) + f(f(x)) \\
f(f(x)+1) +f(x)f(1) = f(f(x)+1) +f(x) \\
f(x)f(1) = f(x)
\end{align*}Suppose that there does not exist $b$ such that $f(b) \neq 0$, then $f(x)=0$ for all real numbers $x$. Otherwise $f(b)f(1) = f(b) \implies f(1) =1 $ as desired.
Claim 3: If $f(a)=0$, then $a=0.$
\Proof: Suppose $f(a)=0$ for some real number $a$. Then $P(a,1)$ gives us:
\begin{align*}
f(f(a)+1)+af(1) = f(a+1) +f(a) \\
f(1) + a = f(a+1) =a+1
\end{align*}On the other hand $P(1,a)$ leads us to the following:
\begin{align*}
f(f(1)+a)+f(a)=f(2a)+f(1) \\
f(a+1) =f(2a)+1 \\
a+1 = f(2a) + 1 \\
f(2a) = a.
\end{align*}Taking $f$ from both sides in the last relation and using (*) leads to:
$$ 0=f(a) = f(f(2a)) =f(2a) = a $$This proves the claim.
To finish the problem, consider $P(x,x-f(x))$:
$$ xf(x-f(x)) =f( (x-f(x)) \cdot (x+1)) $$Setting $x=-1$ gives us:
$$ -f(-1-f(-1)) = f( (1-f(-1)) \cdot 0 ) = f(0) = 0 $$From Claim 3 we obtain that $-1-f(-1)=0 \implies f(-1)=-1$. Now looking at $P(-1,y)$ and replacing $y$ by $y+1$, we get that:
$$ f(y-1) =f(y)-1 \implies f(y+1) = f(y)+1$$On the other hand, $P(x,1)$, the previous relation and (*) give us the following:
\begin{align*}
f(f(x)+1) +x = f(x+1) +f(x) \\
f(f(x)) +1 + x =f(x)+1 +f(x) \\
f(x)+x = 2f(x) \\
f(x) = x
\end{align*}Thus, the only possible functions that satisfy the given relation are $f(x)=x$ and $f(x)=0$. It is easy to check that they indeed work.
tadpoleloop
12.11.2023 00:01
I think maybe I can tidy it up a bit.
Let $P(x,y)$ be the assertion $f(f(x)+y)+xf(y)=f(xy+y)+f(x)$
Clearly $\boxed{f\equiv 0}$ is a solution, so let us assume that $f$ is not identically zero.
Let $a = f(0)$
Then $P(0,y)\implies f(y+a) = f(y) + a$
So $P(a,0)\implies f(f(a)) + a^2 = a + f(a)\implies a^2 = 0 \implies a=0$
$P(x,0)\implies f(f(x)) = f(x)$
$P(f(x),1)\implies f(1)=1$ because $f$ is not zero everywhere.
Let $c = f(-1)$
$P(-1, y)\implies f(y+c) = f(y) + c$
$P(x,c)\implies f(f(x)) + c + cx = f(cx) + c + f(x)\implies f(cx) = cx$
So if $c\ne 0$ then we have the $f$ must be the identity.
Let us assume that $c=0$
$P(1,-1)\implies f(-2) = -1$
But then $-1 = f(-2) = f(f(-2)) = f(-1) = 0$ contradiction.
Thus, $f(-1)\ne 0$ and it follows that $\boxed{f(x) = x}$ is the only other solution.
eulerleonhardfan
12.11.2023 10:42
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.
megarnie
12.11.2023 17:38
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.
Cali.Math
15.01.2024 13:18
We uploaded our solution https://calimath.org/pdf/BalticWay2023-4.pdf on youtube https://youtu.be/TPsMv_clprQ.
solasky
06.06.2024 09:25
We claim that the two solutions are $f \equiv 0$ and $f(x) \equiv x$, which we confirm are solutions. Let $A(x, y)$ denote the given assertion, and let $a := f(0)$. Then, $A(0, x)$ gives us $f(x + a) = f(x) + a$. Using this, we can deduce that
\[ f(a) = f(0) + a = 2a \implies f(f(a)) = f(2a) = f(a) + a = 3a. \]So, $A(a, 0)$ gives us
\[ f(f(a)) + a^2 = f(0) + f(a) \implies a = 0. \]Now, $A(x, 0)$ gives us $f(f(x)) = f(x)$, labelled (1). Then, we can force $f(x) + y$ and $xy + y$ to cancel with $A(f(x), 1)$, which using (1) gives us
\[ f(f(x) + 1) + f(x)f(1) = f(f(x) + 1) + f(x) \implies f \equiv 0 \text{ or } f(1) = 1. \]Since $f \equiv 0$ is one of our claimed solutions, assume otherwise that $f(1) = 1$. $A(1, -1)$ gives us $f(-1) = f(-2) + 1$. Using this and (1) to simplify $A(-2, 1)$ we get that
\[ f(f(-2) + 1) - 2 = f(-1) + f(-2) \implies f(-1) + f(-2) = f(-1) - 2. \]So, $f(-2) = -2$, which in turn gives us $f(-1) = -1$. $A(-1, x + 1)$ gives us $f(x + 1) = f(x) + 1$. By induction, this becomes
\[ f(x + k) = f(x) + k \, \text{ and } f(k) = k \quad (2)\]for all integers $k$. Now, for any integer $k$, using (2) on $A(k - 1, x)$ gives us
\[ (k - 1) + f(x) + (k - 1)f(x) = f(kx) + (k - 1) \implies kf(x) = f(kx). \quad (3)\]Finally, for any integer $k$, using (1), (2), and (3) on $A(x, k)$ gives us
\[ f(x) + k + kx = k(f(x) + 1) + f(x) \implies f(x) \equiv x. \]So, no solutions exist other than $f(x) \equiv x$ and $f \equiv 0$.
MathLuis
28.07.2024 17:37
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