Team Selection Test

$f(x)=2x$, $x\in \mathbb{R}$ is the only solution. Assume that there exists a real number $c\ne 0$ such that $c=f(z)-2z$ for some $z$. Let $P(x,y)$ be the assertion of the given equation. $$P(f(y)-2y,y)\to (f(y)-2y)f(f(y)-2y)=0,\forall y.$$ This shows that $f(c)=0$. Since $-2c=f(c)-2c\ne 0$, we also obtain that $f(-2c)=0$. $$\begin{array}{rl}P(x,c)& \to f(c^2)=f((x+c{)}^2)-xf(x),\\ P(x,-2c)& \to f(4c^2)=f((x-2c{)}^2)-xf(x),\\ P(c,c)& \to f(c^2)=f(4c^2).\end{array}$$ Therefore, we obtain that $f((x+c{)}^2)=f((x-2c{)}^2)$. (1) $$\begin{array}{rl}P(x-2c,c)& \to f(c^2)=f((x-c{)}^2)-(x-2c)f(x-2c),\\ P(x+c,-2c)& \to f(4c^2)=f((x-c{)}^2)-(x+c)f(x+c).\end{array}$$ Hence, we conclude that $(x-2c)f(x-2c)=(x+c)f(x+c)$. (2) Now for all $x$ and $y$ $$P(x+c,y)\to f(xf(y)+y^2+cf(y))=f((x+y+c{)}^2)-(x+c)f(x+c)$$ $$P(x-2c,y)\to f(xf(y)+y^2-2cf(y))=f((x+y-2c{)}^2)-(x-2c)f(x-2c)$$ Using (1) and (2), we get $$f(xf(y)+y^2+cf(y))=f(xf(y)+y^2-2cf(y)).$$ Since the function is surjective, for any two real numbers $u\ne v$, there exist two real numbers $x,y$ such that $$f(y)=\frac{u-v}{3c}\ne 0,x=\frac{u-y^2-cf(y)}{f(y)}$$ This means that $f(u)=f(v)$ for any $u\ne v$, and $f$ is constant which is not possible since it is surjective. Therefore such a $c$ does not exist and $f(x)=2x$, $x\in \mathbb{R}$. This function satisfies the equation: $$ f(xf(y) + y^2) = 4xy + 2y^2 = 2(x+y)^2 - 2x^2 = f((x+y)^2) - x f(x).