Find all functions $f$ from real numbers to real numbers such that $$2f((x+y)^2)=f(x+y)+(f(x))^2+(4y-1)f(x)-2y+4y^2$$holds for all real numbers $x$ and $y$.
Problem
Source: 2024 Taiwan Mathematics Olympiad
Tags: algebra, functional equation, Taiwan
Tintarn
05.02.2024 02:57
Put $x=1, y=0$ to get $f(1)^2=2f(1)$ and hence $f(1)=0$ or $f(1)=2$.
Just put $y=1-x$ to get $f(x)^2+(3-4x)f(x)+4x^2-6x+2-f(1)=0$.
If $f(1)=0$, this means that $0=f(x)^2+(3-4x)f(x)+4x^2-6x+2=(f(x)+1-2x)(f(x)+2-2x)$ and hence $f(x) \in \{(2x-1,2x-2\}$ for all $x$.
For $x,y$ integers, by considering parity we see that $f(x+y)$ is even and hence $f(n)=2n-2$ for all $n \in \mathbb{Z}$.
But now choosing $x=y=2$, we get a contradiction.
If $f(1)=2$, we get $0=f(x)^2+(3-4x)f(x)+4x^2-6x=(f(x)+3-2x)(f(x)-2x)$ and hence $f(x) \in \{2x,2x-3\}$ for all $x$.
As above, we then deduce by parity that $f(n)=2n$ for all integers $n$ and then plugging in $y=2-x$ we get that $f(x)=2x$ for all $x$ which is indeed a solution.
Hence the only solution is $f(x)=2x$.
Korean_fish_Kaohsiung
05.02.2024 08:04
Tintarn wrote:
Put $x=1, y=0$ to get $f(1)^2=2f(1)$ and hence $f(1)=0$ or $f(1)=2$.
Just put $y=1-x$ to get $f(x)^2+(3-4x)f(x)+4x^2-6x+2-f(1)=0$.
If $f(1)=0$, this means that $0=f(x)^2+(3-4x)f(x)+4x^2-6x+2=(f(x)+1-2x)(f(x)+2-2x)$ and hence $f(x) \in \{(2x-1,2x-2\}$ for all $x$.
For $x,y$ integers, by considering parity we see that $f(x+y)$ is even and hence $f(n)=2n-2$ for all $n \in \mathbb{Z}$.
But now choosing $y=n-x$ for $n$ sufficiently large, we get that $f(x)=2x-2$ for all $x$ which is indeed a solution.
If $f(1)=2$, this means that $0=f(x)^2+(3-4x)f(x)+4x^2-6x=(f(x)+3-2x)(f(x)-2x)$ and hence $f(x) \in \{2x,2x-3\}$ for all $x$.
As above, we then first deduce that $f(n)=2n$ for all integers $n$ and then $f(x)=2x$ for all $x \in \mathbb{R}$.
Hence the two solutions are $f(x)=2x$ and $f(x)=2x-2$.
Umm $f(x)=2x-2$ isn't a solution, plug it in and it fails.
Tintarn
05.02.2024 13:29
Oops, thanks for pointing that out, that was a bit unexpected.
Taha-R
22.02.2024 23:46
Answer : $ f(x) = 2x $ for all $ x \in \mathbb{R} $. Solution : Let $ P(x,y) $ denote the given assertion. $ P(0,0) \Rightarrow f(0) = 0 $ or $ f(0) = 2 $. Case 1 : $ f(0) = 0 $. $ P(x,0) \Rightarrow 2f(x^2) = f(x)^2. (*) $ $ P(0,x) $ and $ (*) \Rightarrow f(x)^2 - f(x) + 2x -4x^2 = 0 $. Consider that $ f(x) = A $ then $ A^2 - A +2x -4x^2 = 0 $. By $ \Delta \Rightarrow A = f(x) = 2x $ or $ - 2x + 1 $. which is the result of the placement, the only answer is $ f(x) = 2x $. Case 2 : $ f(0) = 2 $. $ P(x,-x) \Rightarrow f(x)^2 - (4x + 1)f(x) + 4x^2 + 2x - 2 = 0 $. Consider that $ f(x) = B $ then $ B^2 - (4x+1)B + 4x^2 + 2x - 2 = 0 $. By $ \Delta \Rightarrow B = f(x) = 2x + 1 $ or $ - 2x - 1 $. which is the result of the placement, none of them apply in the equation. As a result, the only answer is $ f(x) = 2x $ for all $ x \in \mathbb{R} $.
sami1618
26.06.2024 22:05
The answer is $f(x)=2x$ which can easily be verified to work. Notice that setting $x=y=0$ we must either have $f(0)=0$ or $f(0)=2$. Case 1: $f(0)=0$ Setting $x=0$ yields $2f(y^2)=f(y)-2y+4y^2$. Thus $2f((x+y)^2)=f(x+y)-2(x+y)+4(x+y)^2$ and we can simplify the assertion to $$4x^2+8xy-2x=(f(x))^2+(4y-1)f(x)$$Deriving with respect to $y$ yields $8x=4f(x)$ or $f(x)=2x$. Case 2: $f(0)=2$ Setting $(x,y)=(0,1)$ gives that $f(1)=12$. Setting $(x,y)=(1,-1)$ gives that $4=-40$, a contradiction.