Find all function $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfied \[f(x+y) + f(x)f(y) = f(xy) + 1 \]$\forall x, y \in \mathbb{R}$
Problem
Source: Indonesian TST for IMO 2023 Stage 1: Test 3 - Algebra
Tags: function, functional equation, algebra
ericxyzhu
28.02.2023 01:32
are you sure this doesn't belong to ISL 2022?
DavyDuf
28.02.2023 03:32
Well, ISL won't be released until next IMO 2023, and this problem is indeed a problem for Indonesian TST So, yeah
Taco12
28.02.2023 03:32
countries use ISL in their TSTs..
VMF-er
28.02.2023 07:16
DavyDuf wrote: Find all function $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfied \[f(x+y) + f(x)f(y) = f(xy) + 1 \]$\forall x, y \in \mathbb{R}$ Let $P(x,y)$ be the assertion $f(x+y)+f(x)f(y)=f(xy)+1$. $P(0,0)\implies f(0)^{2}=1$.
$P(x,0)\implies f(x)=1,\quad\forall x\in\mathbb{R}$, which indeed fits.
$P(2,2)\implies f(2)^{2}=1$
f(2)=-1$P(1,1)\implies f(1)^{2}-f(1)-2=0$
f(1)=2$P(x,1)\implies f(x+1)+f(x)=1,\quad\forall x\in\mathbb{R}\implies f(x+2)=f(x),\quad\forall x\in\mathbb{R}$
$P(x,y+2)-P(x,y)\implies f(xy+2x)=f(xy),\quad\forall x,y\in\mathbb{R}$
$x:=\frac{1}{2},y:=2\implies f(2)=f(1)$, contradiction.
f(1)=-1$P(x,1)\implies f(x+1)=2f(x)+1,\quad\forall x\in\mathbb{R}$
$P(x,y+1)-2P(x,y)\implies f(x)=f(xy+x)-2f(xy)-2,\quad\forall x,y\in\mathbb{R}$
$y:=-1\implies f(x)+2f(-x)=-3,\quad\forall x\in\mathbb{R}$
$x:=-x\implies f(-x)+2f(x)=-3,\quad\forall x\in\mathbb{R}$
$\implies f(x)=-1,\quad\forall x\in\mathbb{R}$, which indeed fits.
f(2)=1$P(1,1)\implies f(1)^{2}-f(1)=0$
f(1)=1$P(-1,1)\implies f(0)=1$, contradiction.
f(1)=0$P(x,1)\implies f(x+1)=f(x)+1,\quad\forall x\in\mathbb{R}$
$P(x,y+1)-P(x,y)\implies f(x)=f(xy+x)-f(xy)-1,\quad\forall x,y\in\mathbb{R}$
$y:=\frac{y}{x}\implies f(x)+f(y)=f(x+y)-1,\quad\forall x\neq 0,y\in\mathbb{R}$.
Since $f(0)=-1$, $f(x)+f(y)=f(x+y)-1,\quad\forall x,y\in\mathbb{R}$
Let $g(x)=f(x)+1,\quad\forall x\in\mathbb{R}$.
Then $g(x+y)=g(x)+g(y),\quad\forall x,y\in\mathbb{R}$.
From $f(x+y)+f(x)f(y)=f(xy)+1$ we have $f(x)+f(y)+f(x)f(y)=f(xy)$, implies $g(xy)=g(x)g(y),\quad\forall x,y\in\mathbb{R}$.
From here, it's easy to prove that $g(x)=x,\quad\forall x\in\mathbb{R}$ and $f(x)=x-1,\quad\forall x\in\mathbb{R}$, which indeed fits.
All solution:
$f(x)=-1,\quad\forall x\in\mathbb{R}$,
$f(x)=1,\quad\forall x\in\mathbb{R}$,
$f(x)=x-1,\quad\forall x\in\mathbb{R}$.