This is my progress till now. Please someone help me out. Let $P(x:y)$ denote the assertion in the above equation. Then \begin{align*} P(x:-f(x))\implies f(x^2-f(x))+f(0)=1\qquad\qquad\cdots\cdots\cdots (1)\end{align*}\begin{align*} P(x:-x^2)\implies f(0)=f(f(x)-x^2)+1-2(f(x)-x^2)\qquad\qquad\cdots\cdots\cdots(2)\end{align*} Pluuging the value of $f(0)$ from $(2)$ in $(1)$ we get $f(x^2-f(x))+f(f(x)-x^2)=2(f(x)-x^2)$ which should be easy to solve.

Abhaysingh2003 wrote: This is my progress till now. Please someone help me out. Let $P(x:y)$ denote the assertion in the above equation. Then \begin{align*} P(x:-f(x))\implies f(x^2-f(x))+f(0)=1\qquad\qquad\cdots\cdots\cdots (1)\end{align*}\begin{align*} P(x:-x^2)\implies f(0)=f(f(x)-x^2)+1-2(f(x)-x^2)\qquad\qquad\cdots\cdots\cdots(2)\end{align*} Pluuging the value of $f(0)$ from $(2)$ in $(1)$ we get $f(x^2-f(x))+f(f(x)-x^2)=2(f(x)-x^2)$ which should be easy to solve. It should be -f(0) for (1)

Let $P(x,y)$ be the assertion in the original equation, $P(y,x^2) \rightarrow f(f(y)+x^2)+1 = f(y^2 + x^2) + 2f(y) + 2x^2 $ $P(x,f(y)) \rightarrow f(f(x)+f(y))+2 = 1 + f(x^2 +f(y)) + 2f(x) + 2f(y) = f(x^2 +y^2) + 2f(x) + 4f(y) + 2x^2 $ From the last equation, we swap $x$ and $y$ and then subtract, we get $$f(x^2 +y^2) + 2f(x) + 4f(y) + 2x^2 = f(f(x)+f(y))+2 = f(y^2 +x^2) + 2f(y) + 4f(x) + 2y^2 $$$$ \Leftrightarrow x^2 - f(x) = y^2 -f(y) = -C$$$$ \Leftrightarrow f(x) ={x^2+C} $$For some real constant $C$, substituting back into $P(x,y)$, we get $ \boxed{C=1}$. Thus the solution is $$\boxed{f(x) = x^2+1 \quad \forall x \in \mathbb{R} }$$@below thank you, I missed that

@above i think your solution is wrong. SomeUser221104 wrote: Let $P(x,y)$ be the assertion in the original equation, $P(y,x^2) \rightarrow f(f(y)+x^2)+1 = f(y^2 + x^2) + 2f(y) + 2x^2 $ $P(x,f(y)) \rightarrow f(f(x)+f(y))+2 = 1 + f(x^2 +f(y)) + 2f(x) + 2f(y) = f(x^2 +y^2) + 2f(x) + 4f(y) + x^2 $ there should be a $2$ in front of the very last $x^2$. You should check your solution on $P(x,y)$ instead of just $P(0,0)$ For a short and simple solution... Note that the assertion is the same as $f(y)+1=f(x^2-f(x)+y)+2y$. Let this be $P(x,y)$. $P(x,0); f(x^2-f(x))=f(0)+1$. $P(x,y^2-f(y)): f(x^2-f(x)+y^2-f(y))=f(0)+2-2f(y^2-f(y))$ and interchanging $x,y$ $f(x)=x^2+c$ for some constant $c$. Checking, $c=1$ so the only solution is $f(x)=x^2+1$

bora_olmez wrote: Abhaysingh2003 wrote: This is my progress till now. Please someone help me out. Let $P(x:y)$ denote the assertion in the above equation. Then \begin{align*} P(x:-f(x))\implies f(x^2-f(x))+f(0)=1\qquad\qquad\cdots\cdots\cdots (1)\end{align*}\begin{align*} P(x:-x^2)\implies f(0)=f(f(x)-x^2)+1-2(f(x)-x^2)\qquad\qquad\cdots\cdots\cdots(2)\end{align*} Pluuging the value of $f(0)$ from $(2)$ in $(1)$ we get $f(x^2-f(x))+f(f(x)-x^2)=2(f(x)-x^2)$ which should be easy to solve. It should be -f(0) for (1) No I think

Abhaysingh2003 wrote: bora_olmez wrote: Abhaysingh2003 wrote: This is my progress till now. Please someone help me out. Let $P(x:y)$ denote the assertion in the above equation. Then \begin{align*} P(x:-f(x))\implies f(x^2-f(x))+f(0)=1\qquad\qquad\cdots\cdots\cdots (1)\end{align*}\begin{align*} P(x:-x^2)\implies f(0)=f(f(x)-x^2)+1-2(f(x)-x^2)\qquad\qquad\cdots\cdots\cdots(2)\end{align*} Pluuging the value of $f(0)$ from $(2)$ in $(1)$ we get $f(x^2-f(x))+f(f(x)-x^2)=2(f(x)-x^2)$ which should be easy to solve. It should be -f(0) for (1) No I think Do you not get $ f(0) + 1 = f(x^2-f(x)) $ ?

@above $f(0)+1=f(x^2-f(x))+2f(0)$. So it just suffices to solve this equation ($f(x^2-f(x))+f(f(x)-x^2)=2(f(x)-x^2)$). Can anyone please help me out? $

No, it is $f(0) + 1 = f(x^2-f(x))$... look at gghx's solution as well if you don't believe me. $P(x,-f(x)$ implies $ f(0) + 1 = f(x^2-f(x)) $, so the last line in your first post should be: $f(x^2-f(x))-f(f(x)-x^2) = 2(x^2+1-f(x))$. This means that it suffices to prove that f is even (as, if the function is even then 0=RHS) ... which can be done using the methods by gghx or SomeUser221104 (by switching x and y in $P(x,f(y))$ or $P(x, y^2-f(y))$ )

Miku3D wrote: Find all such functions $f:\mathbb{R} \to \mathbb{R}$ that for any real $x,y$ the following equation is true. $$f(f(x)+y)+1=f(x^2+y)+2f(x)+2y$$ let $p(x,y)=f(f(x)+y)+1=f(x^2+y)+2f(x)+2y$ we prove if $f(x)=f(y)$ then $x^2=y^2$ we have from $p(x,z)-p(y,z)$ that $x^2-y^2=t$ then $f(z+t)=f(z)$ giving $t=0$ by checking $p(u,z+t)-p(u,z)$ so $f$ is kinda injective , we have from $p(x,\frac{1}{2}-f(x))$ that $f(\frac{1}{2})=f(x^2-f(x)+ \frac{1}{2})$ giving $f(x)=x^2+1$ or $f(x)=x^2$ we prove $f(x)=x^2$ is not pissible by checking $p(x,y)$ we get $1=x^2+2y$ a contradiction so $f(x)=x^2+1$

Let $P(x,y)$ be the assertion as always. $P(x,f(y))\implies f(f(x)+f(y))+1=f(x^2+f(y))+2f(x)+2f(y)$ $P(y,f(x))\implies f(f(x)+f(y))+1=f(y^2+f(x))+2f(x)+2f(y)$ Hence, subtracting those, we obtain $f(x^2+f(y))=f(y^2+f(x))$. $P(x,y^2)\implies f(f(x)+y^2)+1=f(x^2+y^2)+2f(x)+2y^2$ $P(x,y^2)\implies f(f(y)+x^2)+1=f(x^2+y^2)+2f(y)+2x^2$ Now, subtracting those two, we have $f(x)-x^2=f(y)-y^2$ for any $x,y\in\mathbb R$. Therefore, we conclude that $f$ is in the form $f(x)=x^2+c$, where $c$ is just a constant. Plugging this back in, we get $\mathcal{LHS}=f(f(x)+y)+1=f(x^2+y+c)+1=x^4+y^2+c^2+2yc+2x^2c+2x^2y+c+1$ $\mathcal{RHS}=f(x^2+y)+2f(x)+2y=x^4+y^2+2x^2y+c+2x^2+2c+2y$ Subtracting and grouping, we get $(c-1)(c-1+2x^2+2y)=0\implies c=1$. Answer. $\boxed{f(x)=x^2+1\forall x\in\mathbb R}.$

The only answer is $f(x)=x^2+1$. Let $P(x,y)$ be the given assertion. From $P(x,-f(x))$ we have that $f(x^2-f(x))=1+f(0)=c$. We define a set $S$ to be the set of all numbers $x$ such that $f(x)=c$. Now let's assume that $card(S)\geq 3$. From $P(y,x^2)$ and $P(x,y^2)$ we have that $f(f(y)+x^2)-f(f(x)+y^2)=2(f(y)-f(x)+x^2-y^2)$. Now let's say that $x,y\in S$ we have that $f(c+x^2)-f(c+y^2)=2(x^2-y^2)$. Under our assumption that $card(S) \geq 3$, we can fix $y$, thus we have that for every $x \in S$ the following $f(c+x^2)=2x^2+b$ holds. Now $P(x,c)$, where $x \in S$, we have that $f(2c)=2x^2+b+4c-1$, but since $card(S) \geq 3$, we easily get a contradiction on the assumption that $card(S) \geq 3$. Thus we have the following cases. 1.$card(S)=1$,this means that $f$ is injective when $f(x)=c$. This means that $x^2-f(x)=C \implies f(x)=x^2-C$, but when plugged in we have that $C=-1$, but we get a contradiction on the injectivity part. 2.$card(S)=2$, this easily implies that if some $b \in S$, then the only other member is $-b \in S$. Thus we have two options $f(x)=x^2-b$ or $f(x)=x^2+b$. Plugging in we get that $f(x)=x^2+1$.

Take y= -f(x) and see f(0) + 1 = f(x^2-f(x)) rest is easy f is constant or f(x)-x^2 constant. But f(x) cannot be constant because this 1=0 a condration. So only case is f(x)=x^2+c , When we write in the equalty we get f(x)=x^2 +1

Which countries contest is this or is it international?

here is the forum aops collection, an international contest it is (in it's 2nd year, in 2020 participated ten countries from Europe, Asia and Africa according to here)

$P(x,y-f(x))\Rightarrow f(x^2-f(x)+y)=f(y)-2y+1$ $P(0,x^2-f(x))\Rightarrow f(x)=x^2+c$, testing gives $\boxed{f(x)=x^2+1}$ which works.

parmenides51 wrote: here is the forum aops collection, an international contest it is (in it's 2nd year, in 2020 participated ten countries from Europe, Asia and Africa according to here) Thanks a lot.

Easy but an excellent problem to remind one how useful bumps in symmetry in FEs can be Miku3D wrote: Find all such functions $f:\mathbb{R} \to \mathbb{R}$ that for any real $x,y$ the following equation is true. $$f(f(x)+y)+1=f(x^2+y)+2f(x)+2y$$ Let $P(x,y)$ be the assertion in the original equation, $P(x,f(y)) \implies f(f(x)+f(y))+1 = f(x^2 +f(y)) + 2f(x) + 2f(y)$ Interchanging $x$ and $y$ in the above equations leads to the conclusion $f(x^2 +f(y)) = f(y^2 +f(x)) $ Now, $P(x,y^2) \implies f(y^2 +f(x)) + 1 = f(x^2+y^2) +2f(x) + 2y^2$ Once again, interchanging $x$ and $y$, and using the equations above, $2f(x) + 2y^2 = 2f(y)+2x^2$ Substituting $y=0$ gives $f(x)=x^2+f(0)$ Finally, Substituting this into the original equation, or using $P(0,0)$ we conclude that $f(0)=1$ Thus the solution is $$\boxed{f(x) = x^2+1 \quad \forall x \in \mathbb{R} }$$

Quote: Find all functions $f:\mathbb{R} \to \mathbb{R}$ satisfying: $$f(f(x)+y)+1=f(x^2+y)+2f(x)+2y,\forall x,y\in \mathbb{R}^{(1)}$$ $(1). P(0,y)\Rightarrow f(y+c)+1=f(y)+2c+2y\Rightarrow f(x+c)=f(x)+2x+2c-1$ (here $f(0)=c$) $(1). P(x,y+c)\Rightarrow f(f(x)+y)+2(f(x)+y)+2c-1+1=f(x^2+y)+2(x^2+y)+2c-1+2f(x)+2(y+c)$ $\Rightarrow f(f(x)+y)+1=f(x^2+y)+2x^2+2y+2c^{(2)},$ from $(1)$ and $(2)$ we have $f(x)=x^2+c$ Subtituting into $(1)$ gives us $\boxed{f(x)=x^2+1,\forall x\in \mathbb{R}}$