Denote by $P(x,y)$ the given assertion. $P(0,y)\implies f(f(y))=y^2+f(0)$ for all $y\in\mathbb{R}.$ Then \begin{align*}f\left(P(x,y)\right) &\implies f(f(x+f(y))) = f(y^2+f(x)) \\ &\implies \left(x+f(y)\right)^2+f(0) = f(y^2)+x^2 \\ &\implies 2xf(y)+f(y)^2+f(0) = f(y^2)\end{align*}Putting $x=y=0$ into this new equation gives $f(0)^2+f(0)=f(0)\implies f(0)=0.$ Now note the following: \begin{align*}P(x^2,f(y)) &\implies f(x^2+f(f(y)))=f(x^2)+f(y)^2 \\ &\implies f(x^2+y^2)=f(x^2)+f(y)^2, \\ f\left(P(f(x),y)\right) &\implies f\left(f(f(x)+f(y))\right)=f\left(f(f(x))+y^2\right) \\ &\implies \left(f(x)+f(y)\right)^2=f(x^2+y^2),\end{align*}from which it follows by transitivity that \begin{align*}f(x^2)+f(y)^2=\left(f(x)+f(y)\right)^2 &\implies f(x^2)=f(x)^2+2f(x)f(y).\end{align*}Earlier we found that $f(y^2)=2xf(y)+f(y)^2+f(0).$ Using $f(0)=0$ and switching $x,y$ gives $f(x^2)=2yf(x)+f(x)^2.$ Combining this with the above gives $2f(x)f(y)=2yf(x)$ for all real $x,y,$ so either $f(x)\equiv0$ or $f(x)\equiv x$ for all $x\in\mathbb{R}.$ Neither is a solution to the original equation, so we can conclude that there are no solutions at all.

parmenides51 wrote: Show that there are no functions $f : R \to R$ satisfying $f(x + f(y)) = f(x) + y^2$ for all real numbers $x$ and $y$ Let $P(x,y)$ be the assertion $f(x+f(y))=f(x)+y^2$ Let $a=f(0)$ $P(0,x)$ $\implies$ $f(f(x))=x^2+a$ $P(f(x)-f(y),y)$ $\implies$ $f(f(x)-f(y))=x^2-y^2+a$ and so $f(x)$ is surjective $P(x,y)$ $\implies$ $f(x+f(y))\ge f(x)$ and, since surjective, $f(z)\ge f(x)$ $\forall x,z$ So $f(x)$ must be constant, which is never a solution. Q.E.D.

Why follow from Quote: $P(f(x)-f(y),y)$ $\implies$ $f(f(x)-f(y))=x^2-y^2+a$ and so $f(x)$ is surjective that the function it is surjective?

Obvious, every real number can be written as $x^2-y^2+a$ for some $x,y\in {\bf R}$.

The definition of the surjectivity it is $f:A\to B$ it is surjective only if $\forall t\in B,\exists z\in A,t=f(z)$. How you connect your observation to this definition?

Take $t$, write $t=x^2-y^2+a$ for some $x,y$, then you also have $t=f(f(x)-f(y))$. Hence $t=f(z)$ when you put $z:=f(x)-f(y)$. It seems quite clear to me.

Yes indeed, thank you.

My solution. Let P(x,y) be $f(x + f(y)) = f(x) + y^2$ . If f(a)=f(b) we have $f(x+f(a)) = f(x+f(b))$, $f(x)+a^2=f(x)+b^2$ implies $a=b$ or $a=-b$. Puting P(0,0) we get $f(f(0))=f(0)$ implies f(0)=0. Let a be any nullpoint. Then P(x,a) gives $f(x)=f(x)+a^2$ . Now we conclude that f has only one nullpoint a=0. Putting P(0,y) gives $f(f(y))=y^2$ and we can conclude that for every nonegative y there exists x such that $f(x)=y$. Let x be any negative number. Then there exists an y such that $f(y)=-x$. Putting the last in $f(x + f(y)) = f(x) + y^2$ we get $f(0)=f(x)+y^2$ implies $f(x)=-y^2$ implies that $f(x)<=0$. If we suppose that for every x<0 $f(x)<0$ using $f(f(y))=y^2$ we get that left side of the last equation is negative, but the right side is nonnegative and we got a contradiction. We conclude that there must exist one negative x such that $f(x)=0$, but that implies x=0 and we got a contradiction which means that such function doesn't exist.

(1),if f(x)≠0 for any x∈R, then by (x,0) we have f(x+f(0))=f(x), so f is periodic. by (0,y) we have f(f(y))=f(0)+y^2, so for any x∈R, by f(f(y))=f(f(y+f(0))) we conclude y^2=(y+f(0))^2, so f(0)=0, contradiction! (2), if there exists u∈R, f(u)=0, then (x,u) shows that u=0, namely, f(0)=0 and if x≠0, then f(x)≠0. by (0,y)we have f(f(y))=y^2, so by (x,f(1)) and f(f(1))=1^2=1 we have f(x+1)=f(x)+f(1)^2>f(x) (for f(1)≠0!!) finally, by (1,1) we have f(1+f(1))=1+f(1), (1+f(1))^2=f(LHS)=f(RHS)=1+f(1) (for the first"=", pay attention that f(f(y))=y^2), and f(1)≠0, so f(1)=-1<f(0) contradiction! by (1) and (2), we know that such f doesn't exist.

Can somebody delete code marks for me? I'm a new user so I can't use latex. [quote=BobThePotato]Denote by $P(x,y)$ the given assertion. $P(0,y)\implies f(f(y))=y^2+f(0)$ for all $y\in\mathbb{R}.$ Then \begin{align*}f\left(P(x,y)\right) &\implies f(f(x+f(y))) = f(y^2+f(x)) \\ &\implies \left(x+f(y)\right)^2+f(0) = f(y^2)+x^2 \\ &\implies 2xf(y)+f(y)^2+f(0) = f(y^2)\end{align*} Putting $x=y=0$ into this new equation gives $f(0)^2+f(0)=f(0)\implies f(0)=0.$[/quote] At this point I think that we can use $2xf(y)+f(y)^2 = f(y^2)$ and take any $a\in\mathbb{R}$ s.t. $f(a) \neq 0$ (if there is no such $a$ then f is constant which is impossible), so we have $x = \frac{f(a^2) - f(a)^2}{2f(a)}$ for all $x\in\mathbb{R}.$ We can see that RHS is constant and LHS isn't which is a contradiction.

kawazmlekiem wrote: Can somebody delete code marks for me? I'm a new user so I can't use latex. BobThePotato wrote: Denote by $P(x,y)$ the given assertion. $P(0,y)\implies f(f(y))=y^2+f(0)$ for all $y\in\mathbb{R}.$ Then \begin{align*}f\left(P(x,y)\right) &\implies f(f(x+f(y))) = f(y^2+f(x)) \\ &\implies \left(x+f(y)\right)^2+f(0) = f(y^2)+x^2 \\ &\implies 2xf(y)+f(y)^2+f(0) = f(y^2)\end{align*}Putting $x=y=0$ into this new equation gives $f(0)^2+f(0)=f(0)\implies f(0)=0.$ At this point I think that we can use $2xf(y)+f(y)^2 = f(y^2)$ and take any $a\in\mathbb{R}$ s.t. $f(a) \neq 0$ (if there is no such $a$ then f is constant which is impossible), so we have $x = \frac{f(a^2) - f(a)^2}{2f(a)}$ for all $x\in\mathbb{R}.$ We can see that RHS is constant and LHS isn't which is a contradiction.

Nice problem $\color{black}\rule{25cm}{1pt}$ Let $P(x,y)$ denote the given assertion. Since we have that $y^2=(-y)^2$, from comparing $P(x,y)$ and $P(x,-y)$ we have that: $$f(x+f(y))=f(x)+y^2=f(x)+(-y)^2=f(x+f(-y))$$ Let's say that there exist some $x$ and $a$ such that we have that $f(x)=f(x+a)$ and $a \neq 0$, then we have that $P(y,x)=P(y,x+a)$ this implies that: $$f(y)+x^2=f(y)+(x+a)^2$$in turn we must have that $a=-2x$ this in turn gives us that $f(x)=f(-x)$, which implies that $f$ is an even function. Thus from this we easily have that $f(x+f(y))=f(f(y)-x)$. Now let's set $x=f(y)$, from the upper relation we have that $f(2f(y))=f(0)$. But from $P(-f(y),y)$ we have that $f(0)=f(-f(y))+y^2=f(f(y))+y^2$, also notice from $P(0,y)$ we have that $f(f(y))=f(0)+y^2$. Thus if we plug in $f(f(y))$ we have that: $$f(0)=f(f(y))+y^2=f(0)+y^2+y^2$$this implies that $y^2=0$, which is a contradiction since $y$ can be any possible value in $\mathbb{R}$. This implies that $a=0$, which in turn would imply that $f$ is an injective function. On the other hand we have that $f(x+f(y))=f(x+f(-y))$, this implies that $f(y)=f(-y)$ which contradicts the injectivity of $f$. Thus we have that no function satisfies the given assertion.

Let $f(0)=a$. From $P(0,0)$ we get $f(a)=a$, so $P(0,a)\implies a=0$. Thus, we see from $P(0,x)$ that $f(f(x))=x^2$, or $f(x^2)=(f(x))^2$ by the associative trick Let $g(x)=f(f(x))=x^2$, then $f(f(f(x)))=f(g(x))=g(f(x))$, so $f(x^2)=(f(x))^2$ . Hence, $f(1)=(f(1))^2$, implying $f(1)=0$ or $1$. In the former case, we get a contradiction as $P(x,1)\implies 0=1$, so we must have $f(1)=1$. Now, $P(x,1)\implies f(x+1)=f(x)+1$, so $f(n)=n$ for all integers $n$ by induction (as $f(0)=0$). However, $P(2020,2021)\implies 2020+2021=2020+2021^2$ which is a contradiction, so there are no solutions to the functional equation.

Let $y=0$. Then $f(x+f(0))=f(x)$, so $f$ is periodic. Now let $x=0$, so $f(f(y))=f(0)+y^2$. Then $f(f(y))$ is periodic, but $f(0)+y^2$ isn't. Therefore contradiction.

P(0,y): f(f(y)) = f(0) + y^2 => [f(0),∞) is part of the image of f P(-f(y),y): f(0) = f(-f(y)) + y^2 => f(-f(y)) = f(0) - y^2 => ( -∞,f(0)] is part of the image of f Therefore f is surjective. Let c be a number with f(c) = 0 P(x,c): f(x) = f(x) + c^2 => c = 0 Therefore f(c) = 0 if and only if c = 0 Let f(y') = y for any y. P(0,y'): f(y) = y'^2 P(x,y'): f(x + y) = f(x) + f(y) f is the additive function for all y , (y + 1)^2 = f(f(y + 1)) = f(f(y) + f(1)) = f(f(y)) + f(f(1)) = y^2 + k where k is a constant => 2y + 1 = k for all y impossible.

Let $P(x,y)$ denote the assertion $f(x + f(y)) = f(x) + y^2$. $P(0,x)$ gives $f(f(x))=x^2+f(0)$ and $P(-f(x),x)$ gives $f(-f(x))=-x^2+f(0)$. So $f$ is surjective. Let $f(a)=0$, then $P(0,a)$ gives $a=0$. So $f$ is odd. Now $P(0,x)$ and $P(0,-x)$ implies $f$ is even. So $f\equiv 0$, which fails.

Let $P(x,y)$ be the given assertion. $P(0,0): f(f(0)) = f(0)$. $P(0,f(0)): f(0)= f(0) + f(0)^2\implies f(0) = 0$. $P(0,x): f(f(x)) = x^2$. $P(x,f(y)) f(x+y^2) = f(x) + f(y)^2$. Setting $x\to -y^2$ here gives \[f(-y^2) + f(y)^2 = 0,\]so $f(x) \le 0 \forall x\le 0$. Fix an $x< 0$. We have $f(f(x)) = x^2 >0$. However, $f(x) \le 0$ implies $f(f(x))\le 0$, contradiction.

Easy one