iflugkiwmu wrote: Find all functions $f \colon \mathbb{R} \to \mathbb{R}$ such that $f ( f ( x ) + y^2) = x - 1 + ( y+1)f(y)$ holds for all real numbers $x, y$. Let $P(x,y)$ denote the given assertion. $P(x,-1)$ implies $f(f(x)+1)=x-1$, hence $f$ is bijective. $P(x,0)$ yields $f(f(x)) = x - 1 + f(0)$. Since $f$ is surjective, we get $f(z+1) = f(z) - f(0)$ for all $z$. In particular, plugging in $z=0$ gives $f(1)=0$. Next, $P(1,x)$ implies $f(x^2) = (x+1)f(x)$. Hence, we get $(1-x)f(-x) = (1+x)f(x)$ for all $x$. Now pick $a \in \mathbb{R}$ with $f(a)=1$. $P(a,x)$ yields $$f(x^2) - f(0) = f(x^2+1) = f(x^2+f(a)) = a-1+(x+1)f(x)=a-1+f(x^2) \Longrightarrow a=1-f(0).$$Hence, we obtain that $f(1-f(0))=1$ and $f(-f(0)) = 1+f(0)$. Now $$1-f(0)^2 = (1-f(0)) f(-f(0)) = (1+f(0)) f(f(0)) = f(0)^2-1 \Longrightarrow f(0)^2=1 \Longrightarrow f(0) \in \{-1,1\}.$$We first consider the case $f(0)=1$. Then $f(f(x))=x$ for all $x$. Hence, $P(f(x),y)$ yields $$f(x+y^2) = f(y^2+f(f(x))) = f(x)-1+(y+1)f(y)=f(x)+f(y^2)-1.$$In particular, $f(y^2)+f(-y^2) = 2$ for all $y$. This implies $f(x)+f(-x)=2$ for all $x$. Therefore, we get $$(1+x)f(x)=(1-x)f(-x)=(1-x)(2-f(x)) \Longrightarrow f(x)=1-x,$$which is indeed a solution. On the other hand, if $f(0)=-1$ then $f(x+1)=f(x)+1$ and $f(f(x))=x-2$ for all $x$. Hence, $P(f(x),y)$ yields $$f(x+y^2-2) = f(x)+f(y^2)-1 \Longrightarrow f(x+y^2)=f(x)+f(y^2)+1.$$As before, we obtain $f(x)+f(-x)=-2$ and, together with $(1-x)f(-x) = (1+x)f(x)$, the solution $f(x)=x-1$. Thus there are two solutions: $\boxed{f(x)=x-1}$ and $\boxed{f(x)=1-x}$.

iflugkiwmu wrote: Find all functions $f \colon \mathbb{R} \to \mathbb{R}$ such that $f ( f ( x ) + y^2) = x - 1 + ( y+1)f(y)$ holds for all real numbers $x, y$. Let $P(x,y)$ be the assertion $f(f(x)+y^2)=x-1+(y+1)f(y)$. $P(x,0)\Rightarrow f(f(x))=x-1+f(0)$, so $f$ is bijective and $f(f(0))=f(0)-1$. $P(x,-1)\Rightarrow f(f(x)+1)=x-1$ Then $f(f(f(x))+1)=f(x)-1$, so $f(x+f(0))=f(x)-1$. Moreover, we have $f(x-1)=f(f(f(x)+1))=f(x)+f(0)$, so $f(x+1)=f(x)-f(0)$. Setting $x=0$ in the last one, we obtain $f(1)=0$. Also, $f(1-f(0))=1$ and $f(-f(0))=1+f(0)$ from the first equation. $P(0,x)\Rightarrow f(x^2)=(x+1)f(x)\Rightarrow (-x+1)f(-x)=(x+1)f(x)$ Setting $x=f(0)$, we have: $$(1-f(0))f(-f(0))=(f(0)+1)f(f(0))\Rightarrow (1-f(0))(1+f(0))=(1+f(0))(f(0)-1)\Rightarrow f(0)\in\{-1,1\}.$$ Case 1: $f(0)=1$ What we previously obtained becomes $f(f(x))=x$ and $f(x+1)=f(x)-1$. $P\left(f^{-1}(x-y^2),y+1\right)-P\left(f^{-1}(x-y^2),y\right)\Rightarrow Q(x,y):f(x+2y)=f(x)+f(y)-y-1$ $Q\left(0,\frac x2\right)\Rightarrow f(x)+\frac x2=f\left(\frac x2\right)$ $Q\left(x,\frac y2\right)\Rightarrow f(x+y)=f(x)+f(y)-1$ Let $g(x)=f(x)-1$, then $g(x+y)=g(x)+g(y)$ and $g(x+1)=g(x)-1$. Also, $g(0)=g(1)-1=0$. $P(x,y)$ becomes $g(g(x)+y^2)=x+y+(y+1)g(y)$ Using additivity, this becomes $g(g(x))+g(y^2)=x+y+(y+1)g(y)$. Setting $x=0$ and $y\mapsto x+y$, we have $g(x^2)=x+(x+1)g(x)$ and $g(x^2+2xy+y^2)=x+y+(x+y+1)g(x+y)$. Then $g(2xy)=xg(y)+yg(x)$, and setting $y=\frac12$ we test to get $g(x)=-x$. Then $\boxed{f(x)=-x+1}$, which works. Case 2: $f(0)=-1$ Then $f(f(x))=x+2$ and $f(x+1)=f(x)+1$. $P\left(f^{-1}(x-y^2),y+1\right)-P\left(f^{-1}(x-y^2),y\right)\Rightarrow R(x,y):f(x+2y)=f(x)+f(y)+y+1$ $R\left(0,\frac x2\right)\Rightarrow f(x)-\frac x2=f\left(\frac x2\right)$ $R\left(x,\frac y2\right)\Rightarrow f(x+y)=f(x)+f(y)+1$ Now $h(x)=f(x)+1$ is additive. Now we have: $$h(h(x)+y^2)=x-y+(y+1)h(y),$$since $h(x+1)=h(x)+1$. Setting $y=0$, we have $h(h(x))=x$, then additivity gives $h(x^2)=-x+(x+1)h(x)$. Then $h(x^2+2xy+y^2)=-x-y+(x+y+1)(h(x)+h(y))$, so: $$h(2xy)=xh(y)+yh(x),$$and as before we test to find $h(x)=x$. Then $\boxed{f(x)=x-1}$, which also works.

iflugkiwmu wrote: Find all functions f : R-->R such that f (f (x) + y^2) = x −1 + (y + 1)f (y) holds for all real numbers x, y Sorry for bad formatting, this is my first long post Let $g(x)=f(x+1)$. The given equation is equivalent to $$g(g(x-1)+y^2-1)=x-1+(y+1)g(y-1)$$Which is equivalent to $$g(g(x)+y^2+2y)=x+(y+2)g(y)$$Let $P(x,y)$ denote this equation. $P(x,-2$ gives that $g$ is bijective. Therefore there must exist a real number $\alpha$ such that $g(\alpha)=0$. $P(\alpha,-1$ gives $g(-1)=\alpha+g(-1)$. Therefore $\alpha=0$ and $g(0)=0$. Using this, $P(x,0)$ gives $g(g(x))=x$. Also, $P(0,y)$ gives $$g(y^2+2y)=(y+2)g(y)$$Now, we can get $P(g(x),y)$ is equivalent to $$g(g(g(x))+y^2+2y)=g(x)+(y+2)g(y)$$$$g(x+y^2+2y)=g(x)+g(y^2+2y)$$Therefore $g(a+b)=g(a)+g(b)$ holds for all values of $a\in\mathbb{R}$ and $b\in\mathbb{R}^+$. This implies $g(-x)+g(x)=g(0)=0$, and therefore $g(-x)=-g(x)$ for all real values of $x$ and $g$ is additive over the real numbers. Finally, combining this with $g(y^2+2y)=(y+2)g(y)$ we get $$g(y^2+2y)=(y+2)g(y)$$$$g(y^2)+2g(y)=yg(y)+2g(y)$$$$g(y^2)=yg(y)$$And setting $y:=g(y)$ is this equation gives $$g(g(y)^2)=g(y)y$$We combine this with $g(y^2)=yg(y)$ to get that $$g(g(y)^2)=g(y^2)$$$$g(y)^2=y^2$$$$g(y)=\pm y$$Suppose there exist $0\neq x \neq y\neq 0$ such that $g(x)=x$ and $g(y)=-y$. This would imply $$\pm_1(x+y^2+2y)=x-y^2-2y$$$$\pm_2(y+x^2+2x)=y-x^2-2x$$Which doesn't hold for any nonzero value of $x,y$ and any election of signs for the $\pm$ (the unique nonzero solution for the first equation is $(x,y)=(x,-2)$ and for the second is $(x,y)=(-2,y)$. Therefore $x=y=-2$, but x and y have to be different and therefore there's no solution) Therefore $g(x)=\pm x$ with the same sign for all $x$. And this gives the solutions $\boxed{f(x)=x-1}$ and $\boxed{f(x)=1-x}$, which clearly work.

@above $g(y)^2=y^2$ does not imply $g(y)=y$ for all $y$ or $g(y)=-y$ for all $y$. There can be a solution such that $g(1)=1$ and $g(2)=-2$.

SerdarBozdag wrote: @above $g(y)^2=y^2$ does not imply $g(y)=y$ for all $y$ or $g(y)=-y$ for all $y$. There can be a solution such that $g(1)=1$ and $g(2)=-2$. Oh, you are right. Thanks for pointing out, I think it's now fixed

$$\pm(x+y^2+2y)=x-y^2-2y$$For +, y=-2

For -, x=0.

SerdarBozdag wrote: $$\pm(x+y^2+2y)=x-y^2-2y$$For +, y=-2 Oh you are right again, i mistakely assumed $y\neq0$ implied $y^2+2y\neq0$. I think its corrected now, thanks again for pointing out (I think your solution for - doesnt work since I assumed $x,y\neq0$)

Let $P(x.y)$ be $f(f(x)+y^2))=x-1+(y+1)f(y)$ $$P(x,0)\Rightarrow f(f(x))=x-1+f(0)$$Then $f$ is a bijection Take $u$ such that $f(u)=0$ $$P(u,0)\Rightarrow f(0)=u-1+f(0)\Rightarrow u=1$$Then $f(1)=0$ $$P(1,y)\Rightarrow f(y^2)=(y+1)f(y)...(1)$$$$(1,-y)\Rightarrow f(y^2)=(-y+1)f(-y)...(2)$$$$P(1-(y+1)f(y),y)\Rightarrow f(1-(y+1)f(y))+y^2=1$$take $y=0$ $$f(1-f(0))=1$$let $v=1-f(0)$ $$P(0,v)\Rightarrow f(f(0)+v^2)=v$$Now applies $f$ to both sides: $$ f(0)+v^2-1+f(0)=1\Rightarrow v^2-2v=0\Rightarrow v=2,0\Rightarrow f(0)=-1,1$$Case 1, $f(0)=-1$: We have $f(f(x))=x-2$ $$P(f(x+1),-1)\Rightarrow f(x)=f(x+1)-1\Rightarrow f(x+1)=f(x)+1$$Now the functional equation becomes that: $$f(f(x)+y^2+1))=x+(y+1)f(y)$$Take $x=-(y+1)f(y)$ then $f(f(-(y+1)f(y))+y^2+1)=0\Rightarrow f(-(y+1)f(y))=-y^2$ by $(1)$ we have $f(-f(y^2))=-y^2$ and applies $f$ and we obtain $-f(y^2)+2=f(-y^2)...(3)$. Now, solving the system equation with $(1),(2),(3)$ we have $f(y)=y-1$ for all real $y$ and that is a solution. Case 2, $f(0)=1$ We have $f(f(x))=x$ $$P(f(x), 1)\Rightarrow f(x+1)=f(x)-1$$$$f(f(x)+y^2)+1=f(f(x)+y^2-1)=x+(y+1)f(y)$$Now take $x=-(y+1)f(y)+1$ $$f(-(y+1)f(y)+1)+y^2-1=0\Rightarrow f(-(y+1)f(y)+2)=-y^2\Rightarrow -(y+1)f(y)+2=f(-y^2)\Rightarrow f(-y^2)+f(y^2)=2$$And one more time, solve the system equation and we have $f(y)=1-y$ for all real $y$.

jasperE3 wrote: ... Case 1: $f(0)=1$ What we previously obtained becomes $f(f(x))=x$ and $f(x+1)=f(x)-1$. $P\left(f^{-1}(x-y^2),y+1\right)-P\left(f^{-1}(x-y^2),y\right)\Rightarrow Q(x,y):f(x+2y)=f(x)+f(y)-y-1$ $Q\left(0,\frac x2\right)\Rightarrow f(x)+\frac x2=f\left(\frac x2\right)$ $Q\left(x,\frac y2\right)\Rightarrow f(x+y)=f(x)+f(y)-1$ Let $g(x)=f(x)-1$, then $g(x+y)=g(x)+g(y)$ ... I think from here there is a shorter way to finish it: $Q(0,x/2) $ yields $g(2x) +x = g(x)$. By additivity, $g(2x)=2g(x)$. Hence, $g(x)=-x$ and $f(x)= 1-x$ (and similarly in the second case).

Gryphos wrote: jasperE3 wrote: ... Case 1: $f(0)=1$ What we previously obtained becomes $f(f(x))=x$ and $f(x+1)=f(x)-1$. $P\left(f^{-1}(x-y^2),y+1\right)-P\left(f^{-1}(x-y^2),y\right)\Rightarrow Q(x,y):f(x+2y)=f(x)+f(y)-y-1$ $Q\left(0,\frac x2\right)\Rightarrow f(x)+\frac x2=f\left(\frac x2\right)$ $Q\left(x,\frac y2\right)\Rightarrow f(x+y)=f(x)+f(y)-1$ Let $g(x)=f(x)-1$, then $g(x+y)=g(x)+g(y)$ ... I think from here there is a shorter way to finish it: $Q(0,x/2) $ yields $g(2x) +x = g(x)$. By additivity, $g(2x)=2g(x)$. Hence, $g(x)=-x$ and $f(x)= 1-x$ (and similarly in the second case). Thank you.

$$(y+1)f(y)=(1-y)f(-y)$$$$f(x+y^2-1)=f(x)+f(y^2)$$$$f(x^2)=ax^2+b$$Remaining easy

Find all functions $f \colon \mathbb{R} \to \mathbb{R}$ such that $f ( f ( x ) + y^2) = x - 1 + ( y+1)f(y)$ holds for all real numbers $x, y$.

jasperE3 wrote: Find all functions $f \colon \mathbb{R} \to \mathbb{R}$ such that $f ( f ( x ) + y^2) = x - 1 + ( y+1)f(y)$ holds for all real numbers $x, y$. Let $P(x,y)$ be the assertion of the problem. It's obvious from this equation that $f$ is both injective and surjective. By $P(1,-1)$ we have there is $t$ such that $f(t)=0$. Then $P(t,0)$ infers that $f(1)=0$. $P(1,y)$ gives that $\boxed{f(y^2) = (y+1)f(y)}$. We name this relation as $A$. Also, by $P(x,0)$ we have $\boxed{f(f(x)) = x-1+f(0)}$. We name this relation as $B$. We know from $A$ and $B$ that $f(y^2+f(x)) = f(y^2) + f(f(x)) - f(0)$. This with $f$ being surjective plus some calculations give that $f$ is semi - additive. i.e. for all $x,y\in \mathbb R$ $$f(x+y)=f(x)+f(y)-f(0).$$Thus we have $f(2x) = 2f(x) - f(0)$ and $f(4x) = 4f(x)-3f(0)$. Plugging $y\rightarrow 2y$ in relation $A$ using these two equations we have $$f(4y^2) = 4yf(y)+4f(y)-3f(0),\;\; (2y+1)f(2y) = (2y+1)(2f(y)-f(0))= 4yf(y)+2f(y)-(2y+1)f(0).$$Then $$ 4yf(y)+4f(y)-3f(0) = 4yf(y)+2f(y)-(2y+1)f(0) \rightarrow f(y) = (1-y)f(0). $$Plugging into $P(x,y)$ we get that $f(0)= \pm 1$ and then answers are $\boxed{f(x)= x-1:\; \forall x}$ and $\boxed{f(x)= 1-x:\; \forall x}$.