$x=0 \Longrightarrow f(f(y)) = f^2(0) + y \Longrightarrow f\ bijective$ .So there is $x_0$ such that $f(x_0)=0$ , so $f(f(y))=y$ $x=t \Longrightarrow f(tf(t)+f(y)) = f^2(t) + y $ ; $ x=f(t) \Longrightarrow f(f(t)t+f(y)) = t^2 + y$ , so $f(t)=t $ or $f(t)=-t$ Suppose that there are $a\neq 0\neq b$ such that $f(a)=a , f(b)=-b)$ , so $ f(a^2-b)=a^2+b$ Two cases are possible:$ a^2+b=f(a^2-b)=a^2-b \Longrightarrow b=0 $ (contradiction) , or , $a^2+b=f(a^2-b)=-a^2+b \Longrightarrow a=0 $ (contradiction) Finally the functional equation has only two solutions :$ f_1(t)=t\ ,\ f_2(t)=-t $

It was also given in our pre TST this year Have a look at the third day problem 1 here

$ f(x.f(x)+f(y))=f(x)^{2}+y$ $ \implies$ $ f(f(y))=f(0)^2+y$ $ y=-f(0)^{2}$ $ \implies$ $ f(f(-f(0)^{2}))=0$ ,$ f(-f(0)^{2})=x_{0}$ $ \implies$ $ f(x_{0})=0$ $ f(x_{0}.f(x_{0})+f(y))=f(x_{0})^{2}+y$ $ \implies$ $ f(f(y))=y$ $ f(y_{1})=f(y_{2})$ $ \implies$ $ y_{1}=y_{2}$ $ \implies$ $ f$ is injective function. $ f(f(y))=f(0)^{2}+y=y$ $ \implies$ $ f(0)=0$. $ f(x.f(x))=f(x)^{2}$ and $ f(f(y))=y$ $ f(x)=x+g(x)$ $ \implies$ $ g(y)+g(f(y))=0$ $ x.(x+g(x))+g(x.f(x))=(x+g(x))^{2}$ $ \implies$ $ x^{2}+x.g(x)+g(x.f(x))=x^{2}+2x.g(x)+g(x)^{2}$ $ \implies$ $ g(x.f(x))=x.g(x)+g(x)^{2}$ $ x=1$ $ \implies$ $ g(f(1))=g(1)+g(1)^{2}$ $ \implies$ $ g(f(1))=-g(1)$ $ \implies$ $ g(1)=0$ or $ g(1)=-2$ $ i)g(1)=0$ $ \implies$ $ f(1)=1$ $ \implies$ $ f(f(y)+1))=y+1$ $ f$ is injective function $ \implies$ $ f(y+1)=f(y)+1$ $ \implies$ $ f(y+n)=f(y)+n$ for $ n\in N$ . $ y=0$ $ \implies$ $ f(n)=n$ for $ n\in N$ $ f(0)=f(-1)+1$ $ \implies$ $ f(-1)=-1$ $ \implies$ $ f(n)=n$ for $ n\in Z$ $ \implies$ we get $ f(x)=x$ $ ii)g(1)=-1$ $ \implies$ $ f(y+1)=f(y)-1$ $ \implies$ $ f(x)=-x$ $ \Box$

trivially,f is bijective. so there exists $x$,$f(x)=1$ then $f(x+f(0))=f(x)$ hence $f(0)=0$ let $x=0,y=x$ then$f(f(x))=x$ let $y=0$,then $f(xf(x))=f^2(x)$ by letting $x=f(x)$,we het $f(xf(x))=f(f(x))=x^2$ hence $f^2(x)=x^2$,so $f(x)=x or -x$ if there exist non-zero x,y,$f(x)=x,f(y)=-y$ then $f(x^2-y)=x^2+y$ yielding $x=0$ or $y=0$ contradiction! hence f can't be piecewice,yielding $f(x)=x$ or $f(x)=-x$.

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Plugging in $x = 0$ we get $f(f(y)) = f(0)^2 + y$. Therefore, there exists a real number $r$ such that $f(r) = 0$ (specifically, $f(-f(0)^2)$). Then letting $x = r$ gives us $f(f(y)) = y$ for all $y \in \mathbb{R}$. This tells us that $f(0) = 0$ when we combine it with $f(f(y)) = f(0)^2 + y$. Now letting $y = 0$ we get $$f(xf(x)) = f(x)^2$$ Replacing $x$ with $f(x)$ gives us $$f(f(x)x) = x^2$$ So therefore $f(x) = \pm x$. Suppose we have $f(a) = a$ where $a \neq 0$. Then I will show that $f(x) = x$ for all $x \in \mathbb{R}$. Letting $x = a$ in the original FE gives us $f(a^2 + f(y)) = a^2 + y$. Letting $y = f(t)$ gives us $f(a^2 + t) = a^2 + f(t)$ and so we can let $t$ be anything we want. Suppose we had $f(a^2 + t) = -a^2 - t$ for some $t$. Then we have $-a^2 - t = a^2 + f(t)$ so then $f(t) = -2a^2 - t$. We therefore must have $f(t) = t$ which tells us that $t = -a^2$. That means the only time that $f(x) = -x$ is when $x = 0$. So now we see that $f(x) = x$ and $f(x) = -x$ are the only solutions.

let $x=0 \implies f(f(y))=f^2(0)+y$ implies $f$ is bijective. $\exists!a$ such that $f(a)=0$, let $x=a \implies f(f(y))=y(2) \rightarrow f^2(0)=0 \rightarrow f(0)=0$. Let $y=0 \implies f(xf(x))=f^2(x)(1)$. Replaced $(1)$ and $(2)$ in the original equation we get $f(xf(x)+f(y))=f(xf(x))+f(f(y))$, let $x=f(x), y=f(y)$ we have $f(x^2+y)=f(x^2)+f(y) \leftrightarrow f(x+y)=f(x)+f(y)(*)$,for $x\geq 0, y\in \mathbb{R}$.From$(*)$ let $y=-x \rightarrow -f(x)=f(-x)$.For $x\leq 0$ we have $f(y)=f(y+x-x)=f(y+x)+f(-x)=f(x+y)-f(x) \leftrightarrow f(x+y)=f(x)+f(y)$,for $x\leq 0,y\in\mathbb{R}$ implies $f(x+y)=f(x)+f(y)$,for $x,y\in\mathbb{R}$. Also from the original equation we have $f(xf(x)+f(y))\geq y\leftrightarrow f(x+y)\geq f(y)$, for $x\geq 0, y\in\mathbb{R} \implies f$ is increasing function. Because f is additive and increasing function implies $f(x)=ax$ for some $a\in\mathbb{R}$.Replaced in the original equation we have $a=1$ or $a=-1$. Thus $f(x)=x$ or $f(x)=-x$ is two functions satisfying the conditions of problem.

putting x = y =0, f(f(0)) = f(0)^2 there exists a t such that , f(t) = 0; setting x = t , y = 0; f(f(0)) = 0; so, f(0) = 0; f(f(x)) = x; setting x = f(x), f(xf(x) + f(y)) = x^2 + y which means that f(x) = x or f(x) = -x case 1: f(1) = 1; putting x = 1 , y = x, f(f(x) + 1) = x + 1 easy checking shows that f(x) = x for all real x; case 2: f(1) = -1; putting x = -1 , y = x; f(f(x) - 1) = x + 1 easy checking shows that f(x) = -x; from 2 cases, f(x) = x for all real x or , f(x) = -x for all real x

@fighter Nice solution, brother. I have another idea. Let $P(x,y)$ be $f(xf(x)+f(y))=f(x)^2+y$.Obviously $f$ is surjective.So let $f(c)=0$,then $P(c,y)\rightarrow f(f(y))=y$.$P(f(x),y)\rightarrow f(xf(x)+f(y))=x^2+y$.Thus $f(x)^2=x^2\Leftrightarrow f(x)=\pm x$. We suppose that ∃$a\neq 0$, ∃$b\neq 0$ s.t. $f(a)=a, f(b)=-b$.$P(a,b)\rightarrow f(a^2-b)=a^2+b$.Then $a^2-b=a^2+b\Leftrightarrow b=0$ or $-(a^2-b)=a^2+b\Leftrightarrow a=0$ which are both absurd. Therefore $\forall x\in \mathbb R:f(x)=x$ or $\forall x\in \mathbb R:f(x)=-x$ which both satisfy the condition.

can we find any solution using injectivity

Ans: $f(x)=x, f(x)=-x$. It is easy to see that these satisfy the equation. Let $P(x,y)$ be the given assertion, $P(x,-f(x)^2)\implies f(xf(x)+f(-f(x)^2))=0$, so there exists $k\in\mathbb{R}$ s.t $f(k)=0$. $P(k,k)\implies f(0)=k$ and $P(0,0)\implies f(f(0))=f(0)^2$, so \[f(f(0))=f(k)=0=f(0)^2\implies f(0)=0.\] Then $P(0,x)\implies f(f(x))=x$, so \[P(x,0)\implies f(xf(x))=f(x)^2\]and replacing $x$ by $f(x)$ gives \[f(xf(x))=f(f(x))^2=x^2,\]thus $f(x)^2=x^2\implies f(x)=\pm x$. Assume there exists $x,y\ne 0$ s.t $f(x)=x, f(y)=-y$, then \[P(x,y)\implies f(x^2-y)=x^2+y\implies x^2-y=x^2+y \ \ \textrm{or} \ \ y-x^2=x^2+y \implies x=0 \ \ \textrm{or} \ \ y=0\]which is a contradiction. Thus, $f(x)=x, -x$ are the only solutions. $\blacksquare$

Let $P(x,y)$ be the assertion of $f(xf(x)+f(y))=f(x)^2+y$ $P(0,x)$: $f(f(x))=f(0)^2+x$ there is a real number x and y for which $f(x)=f(y)$ Let $x=f(x) $and $y=f(y)$ $f(f(x))=f(f(y))$ $f(0)^2+x=f(0)^2+y$ x=y thus it is injective. Again let $f(0)^2=c$ and set x=0 and y=y-c $f(f(y-c))=y$ f is surjective too.thus there exist a real number t such that $f(t)=0$ x=t,y=0 $f(f(0))=0$ $P(0,0)$ : $f(0)=0$ $P(0,x) \rightarrow$ $f(f(x))=x$ $P(x,0) \rightarrow$ $f(xf(x))=f(x)^2$ set $x=f(x)$ we get $f(x)=x or -x$ which are solution indeed. Is it right?

RaMlaF wrote: Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $\forall x,y \in \mathbb{R}$ we have \[f(xf(x)+f(y)) = f(x)^2 + y\] Let $P(x,y)$ the assertion of the given FE. $P(0,x)$ $$f(f(x))=f(0)^2+x \implies f \; \text{involution and bijective}$$Hence exists a $c$ such that $f(c)=0$. $P(c,x)$ $$f(f(x))=x \implies f(0)^2=0 \implies f(0)=0$$$P(x,0)$ $$f(xf(x))=f(x)^2$$$P(f(x),y)$ $$f(xf(x)+f(y))=x^2+y$$Let compare this new equation with the older one. $P(x,y)-P(f(x),y)$ $$f(x)^2=x^2 \implies f(x)=x \; \text{or} \; f(x)=-x$$Assume that exists $a,b \ne 0$ such that $f(a)=a$ and $f(b)=-b$. Note that $f$ is odd since $f(f(-x))=-x \implies f(x)=-f(-x)$ becuase if $f$ is even then $x=-x$ and that is not possible. $P(a,b)$ $$\pm(a^2-b)=a^2+b \implies a=0 \; \text{or} \; b=0$$Contradiction!! Then the solutuons are: $\boxed{f(x)=x \; \forall x \in \mathbb R}$ $\boxed{f(x)=-x \; \forall x \in \mathbb R}$ Thus we are done

Let $P(x,y)$ denote the given assertion. $P(0,y)$ gives that $f(f(y)) = f(0)^2 + y$ and so $f$ is injective and surjective. Let $z$ be the unique number s.t $f(z) = 0$. Then, $P(z,y)$ gives that $f(f(y)) = y$ and so $f(0) = 0$ Also, $P(x,0)$ gives that $f(xf(x)) = f(x)^2$ Now, $P(f(x),f(y))$ gives that $f(xf(x) + y) = x^2 + f(y)$. Putting $y = 0$, we get $f(xf(x)) = x^2$ and so $f(x)^2 = x^2 \implies f(x) = \pm x$ for all $x$. Suppose $f(a) = a$ and $f(b) = -b$ and $a, b \neq 0$. Then, $P(a,b)$ gives that $f(a^2 - b) = a^2 + b$. But the RHS is not $\pm$ of the LHS for any $a,b$ and so this is impossible. So, the only solutions are $f(x) = x$ and $f(x) = -x$ @2below - Yeah that works too

My solution is way too easy, kindly check My idea is same till we prove $f(f(y))=y$, after that replace $x$ by $f(x)$, LHS remains the same, compare RHS to get $f(x)^2=x^2$ and then last step same as @above just posted.

L567 wrote: Now, $P(f(x),f(y))$ gives that $f(xf(x) + y) = x^2 + f(y)$. Putting $y = 0$, we get $f(xf(x)) = x^2$ and so $f(x)^2 = x^2 \implies f(x) = \pm x$ for all $x$. Instead of $P(f(x),f(y))$ you could have done $P(f(x),y)$ the LHS would be exactly the same (as I have told in the post above) then and then there won't be any need to substitute $y=0$

Letting $f(0)=k$ we get that $$f(f(x))=k^2+x \implies f \text{is bijective}$$Take $a$ such that $f(a)=0$ and we get upon putting $x=a$ that $$f(f(x))= x \implies k=0$$Now $f(xf(x)+f(y))$ can also be written as $f(f(x)f(f(x))+f(y)$ .Hence we get that $$f(x)^2+y=x^2+y \implies \forall x , f(x) \in \{x,-x\}$$The point wise trap can be easily sorted out by taking $f(x)=x$ and $f(y)=-y$ (both nonzero) but we will get that atleast one of them must be zero.Contradiction. So $$f(x) \equiv x$$or $$f(x) \equiv -x$$

Nobody mentioned that this is Balkan 1997 and 2000.