syk0526 wrote: Find all functions $ f : \mathbb{R} \rightarrow \mathbb{R} $ such that \[ f( xf(x) + 2y) = f(x^2)+f(y)+x+y-1 \] holds for all $ x, y \in \mathbb{R}$. Let $P(x,y)$ be the assertion $f(xf(x)+2y)=f(x^2)+f(y)+x+y-1$ $P(0,0)$ $\implies$ $f(0)=1$ $P(x,-xf(x))$ $\implies$ $f(x^2)=xf(x)-x+1$ $P(0,x)$ $\implies$ $f(2x)=f(x)+x$ From there we'll compute two expressions for $f(4x^2)$ : $f(2x)=f(x)+x$ $\implies$ $f(4x)=f(x)+3x$ $\implies$ $f(4x^2)=f(x^2)+3x^2$ $=xf(x)+3x^2-x+1$ $f(x^2)=xf(x)-x+1$ $\implies$ $f(4x^2)=2xf(2x)-2x+1$ $=2xf(x)+2x^2-2x+1$ So $xf(x)+3x^2-x+1=2xf(x)+2x^2-2x+1$ and so $x(f(x)-x-1)=0$ and so $f(x)=x+1$ $\forall x\ne 0$, still true when $x=0$ And so $\boxed{f(x)=x+1\text{ }\forall x}$ which indeed is a solution.

Taking $x=y=0$, we derive $f(0)=1$. Taking $y=0$, gives $f(xf(x))=f(x^2)+x$. Now taking $y=-xf(x)$ yields \begin{align}f(x^2)=xf(x)-x+1.\end{align}Therefore, $f(xf(x))=xf(x)+1$. Swapping the sign on $x$ in (1) implies $f(x)+f(-x)=2$ for all $x\neq 0$. But this holds for $x=0$, by inspection. Now note \begin{align*} f(xf(x))&=f(-xf(-x)+2x)\\ &= f(x^2)+f(x)-1\\ &= xf(x)+f(x)-x \end{align*}But using the fact that $f(xf(x))=xf(x)+1$, we obtain the solution $\boxed{f=x+1}$.

Use the transformation $g(x)=f(x)-1$ so our functional equation is equivalent to $$g(xg(x)+x+2y)=g(x^2)+g(y)+x+y.$$Let $P(x,y)$ be the assertion into the functional equation in $g$. $P(0,0)\implies g(0)=0$. $P(x,0)\implies$\begin{align} g(xg(x)+x)=g(x^2)+x.\end{align}$P(x,-xg(x)-x)\implies g(x^2)=xg(x).$ Plugging $x=x$ and $x=-x$ in the previous equation gives $$-xg(-x)=g(x^2)=xg(x)\implies x(g(x)+g(-x))=0.$$For $x\neq 0$, we have $g(x)=-g(-x)$ and this is satisfied for $x=0$ so $g$ is odd. $P(-x,x)\implies g(-xg(-x)+x)=g(x^2)+g(x)$. Using the fact that $g(-x)=-g(x)$ from $g$ being odd, we have $g(xg(x)+x)-g(x^2)=g(x)$. But, by $(1)$, we have that $x=g(xg(x)+x)-g(x^2)$. Hence, $g(x)=x$ so $\boxed{f(x)=x+1}$ and we can easily check that it is a solution.$\blacksquare$

syk0526 wrote: Find all functions $ f : \mathbb{R} \rightarrow \mathbb{R} $ such that \[ f( xf(x) + 2y) = f(x^2)+f(y)+x+y-1 \]holds for all $ x, y \in \mathbb{R}$. $$f(xf(x)+2y)=f(x^2)+f(y)+x+y-1$$$P(0,0)$$\implies$$f(0)=1$ $P(0,x)$$\implies$$f(2x)=f(x)+x$ $$f(4x^2)=f(x^2)+3x^2$$$P(x,0)$$\implies$$f(xf(x))=f(x^2)+x$ $$f(xf(x)+2y)=f(xf(x))+f(2y)-1$$$$f(xf(x)+y)=f(xf(x))+f(y)-1$$$y \implies$ $xf(x)$ $\implies$ $f(2xf(x))=2f(xf(x))-1=xf(x)+f(xf(x))$ $$f(xf(x))=xf(x)+1=f(x^2)+x$$$x \implies 2x$ $\implies$$f(4x^2)+2x=2xf(2x)+1$ $$xf(x)+1-x+3x^2+2x=2x^2+2xf(x)+1$$$$\boxed {f(x)=x+1}$$

$$P(0,0) \implies f(0)=1$$$$P(x,0) \implies f(xf(x))=f(x^2)+x \qquad (1)$$$$P(x,-xf(x)) \implies f(x^2)=xf(x)-x+1 \qquad (2)$$$$f(xf(x))=xf(x)+1 \qquad (3)$$$$\text{ replacing x with -x in 2} \implies f(x)+f(-x)=2 \qquad (4)$$$$P(-x,x) \implies f(-xf(-x)+2x)=f(x^2)+f(x)-1 \qquad (5)$$$$\text{ By (4) } \implies f(xf(x))=f(x(2-f(-x))=f(-xf(-x)+2x) \qquad (6)$$$$\text{By (5),(6),(2)} \implies f(xf(x))=f(x^2)+f(x)-1 = xf(x)+f(x)-x \qquad (7)$$$$\text{By (3),(7)} \implies \boxed{f(x)=x+1} \qquad \blacksquare$$