Find $f:\mathbb{R} \to \mathbb{R}$ which satisfies $f(x)^2+2f(x)f(y)+y^2=\big(x-f(y)\big)^2+4f(x)y$.

Question

Find $f:\mathbb{R} \to \mathbb{R}$ which satisfies $f(x)^2+2f(x)f(y)+y^2=\big(x-f(y)\big)^2+4f(x)y$.

Let $P(x, y): f(x)^2+2f(x)f(y)+y^2=\big(x-f(y)\big)^2+4f(x)y.$

$P(0, 0): f(0)^2-2f(0)f(0)=\big(0-f(0)\big)^2.$

$\therefore -f(0)^2=f(0)^2, f(0)=0.$

$P(x, 0): f(x)^2=x^2.$

$\therefore x^2+2f(x)f(y)+y^2=x^2-2xf(y)+y^2+4f(x)y.$

$\Rightarrow 2f(y)(f(x)+x)=4f(x)y, \ f(y)(f(x)+x)=2f(x)y.$

How can I disprove that $f(x)= \begin{cases} x & \text{if ****} \\ -x & \text{if ****}\end{cases}$?

Answer 1

You have shown that $\lvert f(x) \rvert = |x|$ and $$ f(y)(f(x)+x)=2f(x)y $$ for all $x$ and $y$. Consider whether $f(1)$ is $1$ or $-1$.

If $f(1) = 1$, then (plugging in $y = 1$) $$ 1 \cdot (f(x) + x) = 2 f(x) \cdot 1, $$ so $f(x) = x$ for all $x$.

If $f(1) = -1$, then (plugging in $x = y = 1$) $$ -1 \cdot (-1 + 1) = 2 \cdot (-1) \cdot 1, $$ which is false.

Therefore, the only function that satisfies the original equation is $f(x) = x$.

Answer 2

If you apply for $P(x,x)$ you get $2f(x)(f(x)-x)=0$

So $f(x)=0$ or $f(x)=x$ but only the latter one is compatible with $f(x)^2=x^2$

Therefore $f(x)=x$ for all $x$.