$y=f(x)$ give $f(2y)=f(0)+(2y)^{2}.$ $y=-f(x)$ give $f(-2y)=f(0)+(2y)^{2}.$ It is easy to chek, that $f(x)=x^{2}+c$ solution. For proof, that equation had not other solution suufficiently to show, for any y exist x, suth that $|y|=|f(x)|$.

i don`t think that the surjectivity of $f$ is very simple. in fact, i think it`s harder than the "offical" solution.

Hello, IM(f) is a serious problem. For example, f(x) = 0 is a solution (different from x^2 + c) -- Patrick

The solution that I like is as follows. Let $g(x)=f(x)-x^2$, then the functional relation becomes $g(g(x)+x^2+y)=g(g(x)+x^2-y)$ for all real $x,y$. Then $g(g(x)+x^2-g(y)-y^2+z)=g(g(x)+x^2+g(y)+y^2-z)=g(g(y)+y^2-g(x)-x^2+z)$, thus $g(z)=g(2*(g(x)+x^2-g(y)-y^2) +z)$ for all real $x,y,z$. If $g(x)+x^2$ is constant, then we get the solution $f(x)=0$ for all real $x$. If $g(x)+x^2$ is not constant, then $g$ is periodic, say with period $T$. Then from the last equation we put $x=y+T$ to get $g(z)=g(2*(2*y*T+T^2)+z)$ for all real $y$ and $z$. But we can choose $y$ at our will to get $g(z)=g(0)$ for all real $z$ and hence $f(x)=x^2+g(0)$ for all real $x$.

why is $g$ periodic ?

If g(x) + x^2 is not constant than there exists a different from b such that a=g(x)+x^2 and b=g(z)+z^2 for some x and z. g(a+y)=g(a-y) (*) and g(b+y)=g(b-y) for all y. Substitute y by a-y-b in the second relation and you obtain g(b+(a-y-b))=g(b-(a-y-b)) <=> g(a-y)=g(y+2*b-a), but from (*) g(a-y)=g(y+a) => g(y+a)=g(y+2*b-a). If we take z = y+a we wil get g(z) = g(z+2*(b-a)) and because z undergoes R => g is periodic of period 2*(b-a).

Hello Maky! maky wrote: why is $g$ periodic ? Because he has $g(z)=g(2(g(x)+x^{2}-g(y)-y^{2})+z)$ for all real x,y,z So, if $g(x)+x^{2}$ is not a constant, it exists x and y such that $g(x)+x^{2}\neq g(y)+y^{2}$ Then, if I call $T=2(g(x)+x^{2}-g(y)-y^{2})$, $T\neq 0$ and for any real z, $g(z)=g(2(g(x)+x^{2}-g(y)-y^{2})+z)$ ==> $g(z)=g(z+T)$ And g is periodic. -- Patrick

Anto wrote: The solution that i like is as follows : Let g(x)=f(x)-x^2 , then the functional relation becomes g(g(x)+x^2+y)=g(g(x)+x^2-y) for all real x,y. Then g(g(x)+x^2-g(y)-y^2+z)=g(g(x)+x^2+g(y)+y^2-z)=g(g(y)+y^2-g(x)-x^2+z) thus g(z)=g(2*(g(x)+x^2-g(y)-y^2) +z) for all real x,y,z. If g(x)+x^2 is constant then we get the solution f(x)=0 for all real x. If g(x)+x^2 is not constant then g is periodic say with period T. Then from the last equation we put x=y+T to get g(z)=g(2*(2*y*T+T^2)+z) for all real y and z. But we can choose y at our will to get g(z)=g(0) for all real z and hence f(x)=x^2+g(0) for all real x. can someone turn this into $LaTeX$ and explain a bit?

all the solutions i know (including mine, in the contest) use the idea that $\mbox{Im}\, f-\mbox{Im}\, f =\mathbb{R}$. after that, it`s pure algebraic manipulations.

maky wrote: all the solutions i know (including mine, in the contest) use the idea that $\mbox{Im}\, f-\mbox{Im}\, f =\mathbb{R}$. after that, it`s pure algebraic manipulations. can you show that?

What is IR?

Anto wrote: The solution that i like is as follows : Let $g(x)=f(x)-x^{2}$ , then the functional relation becomes \[g\left(g(x)+x^{2}+y\right)=g\left(g(x)+x^{2}-y\right)\qquad \forall \text{ real }x,y\]Then \[g\left(g(x)+x^{2}-g(y)-y^{2}+z\right)=g\left(g(x)+x^{2}+g(y)+y^{2}-z\right)=g\left(g(y)+y^{2}-g(x)-x^{2}+z\right)\] thus \[g(z)=g\left(2(g(x)+x^{2}-g(y)-y^{2}\right)+z)\qquad \forall\text{ real }x,y,z\]If $g(x)+x^{2}$ is constant then we get the solution $f(x)=0$ for all real $x$. If $g(x)+x^{2}$ is not constant then $g$ is periodic say with period $T$. Then from the last equation we put $x=y+T$ to get \[g(z)=g\left(2\cdot(2yT+T^{2})+z\right)\qquad \forall \text{ real }y,z\]But we can choose $y$ at our will to get \[g(z)=g(0)\qquad\forall \text{ real }z\] and hence \[\boxed{f(x)=x^{2}+g(0) \qquad\forall \text{ real }x}\] IR is pure imaginaries.

In problem IR is $\mathbb{R}$ = set of all real numbers.

i will post here my solution. i gave it in the contest as well the solutions are $f\equiv 0$ and $f(x)=x^{2}+a$, with $a$ real. obviously $f\equiv 0$ satisfies the equation, so i will choose an $x_{0}$ now such that $f(x_{0})\neq 0$. i first claim that any real number can be written as $f(u)-f(v)$, with $u,v$ reals. denote $f(x_{0})$ with $t\neq 0$. then, putting in the equation, it follows that $f(t+y)-f(t-y)=4ty$. since $t\neq 0$ and $y$ is an arbitrary real number, it follows that any real $d$ can be written as $f(u)-f(v)$, with $u,v$ reals (take $y=d/4t$ above). this proves my claim above. now, let`s see that $f\big(f(x)+f(y)+z\big)=f\big(f(x)-f(y)-z\big)+4f(x)\big(f(y)+z\big)$ and $f\big(f(y)+f(x)+z\big)=f\big(f(y)-f(x)-z\big)+4f(y)\big(f(x)+z\big)$ (the relations are deduced from the hypothesis) for all reals $x,y,z$. i will denote by $d$ the difference $f(x)-f(y)$. substracting the above relations will give me $f(d-z)-f(-d-z)+4zd=0$. now, let`s see that this is true for all reals $d,z$, because $z$ was chosen arbitrary and $d=f(x)-f(y)$ can be choosen arbitrary because of the first claim. now it`s easy. just take $d=z=-x/2$ and it gets that $f(0)-f(x)+x^{2}=0$, or $f(x)=x^{2}+f(0)$. this obviously satisfies the relation, and this ends the proof.

First $ f(x)=0$ is solution. Now suppose $f(x) \not=0$ for some $x$. We can see that any $y\in R$ can be represented by $f(a)-f(b)$ and also $2(f(a)-f(b))$ for some reals $a, b$ (1). Now placing $f(x)=y$ gives $f(2f(x))=f(0)+4f(x)^2$. Put $y=f(x)-2f(y)$ and we get $f(2(f(x)-f(y)))=f(2f(y))+4f(x)^2-8f(x)f(y)$ $=f(0)+4f(y)^2+4f(x)^2-8f(x)f(y)$ $=f(0)+4(f(x)-f(y))^2$. Because of (1) we see that all solutions other than $ f(x)=0$ are $f(x)=x^2+f(0)$

From which source did you find this question. It is International Zhautykov olympiad 2011.But you wrote it at 2007 How?

ShahinBJK wrote: From which source did you find this question. It is International Zhautykov olympiad 2011.But you wrote it at 2007 How? It is Balkan MO 2007 second problem. See here: http://www.artofproblemsolving.com/Forum/resources.php?c=1&cid=19&year=2007&sid=e95c823acb9d986b9856ea76c176582b

Not quite. Swapping $x$ and $y$, the Zhautykov writes as $f(f(x) + y) = f(y - f(x)) + 4f(x)y$ (the difference is, at the Balkan MO, the start of RHS was $f(f(x) - y)$). But, of course, the method is identical, so the problem is a spoof. See also my comment posted at the Zhautykov link http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2148157&sid=3a2481060d2b70418b0728a796017e01#p2148157.

mavropnevma wrote: Not quite. Swapping $x$ and $y$, the Zhautykov writes as $f(f(x) + y) = f(y - f(x)) + 4f(x)y$ (the difference is, at the Balkan MO, the start of RHS was $f(f(x) - y)$). But, of course, the method is identical, so the problem is a spoof. See also my comment posted at the Zhautykov link http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2148157&sid=3a2481060d2b70418b0728a796017e01#p2148157. yes i didn`t see the difference but $f:even$ can be easily found.

IR means irrational?

No,it is the set of real numbers.

Does this work? $f \equiv 0$ is trivial. Assume there exists $x_0$ such that $f(x_0) \neq 0$. The equation is equivalent to $f(x+y)=f(x-y)+4xy$ for all $x \in \mbox{Im}\,f$, $y \in \mathbb{R}$. (1) Since $y$ can range over all real numbers, sub $x=x_0$ gives $\mbox{Im}\,f - \mbox{Im}\,f = \mathbb{R}$. Let $f(x)-x^2=g(x)$. Then, (1) is equivalent to $g(x+y)=g(x-y)$ or $g((x-z)+y+z)=g((x-z)-y+z)$ for all $x,z \in \mbox{Im}\,f$, $y \in \mathbb{R}$. Since $\mbox{Im}\,f - \mbox{Im}\,f = \mathbb{R}$, we get $g(w+y+z)=g(w-y+z)$ for all $z \in \mbox{Im}\,f$, $y,w \in \mathbb{R}$. Let $y= \frac{u-v}{2}$ and $w=\frac{u+v-2z}{2}$ for arbitrary real numbers $u, v$. Then, $g(u)=g(v)$ for all $u,v \in \mathbb{R}$ i.e. $g$ is a constant i.e. $f(x)=x^2+c$ for some constant $c$, as required.

Hi MathPanda1, I am not sure why if you fix a real z in Im(f), then x-z can represent all reals? If not, then w is only the set Im(f)-z, which is not R. (since Im(f) is not necessarily R)

Redacted...

Im(f) here i mean image of f, not the imaginary part of f. I am sorry for the confusion

Anto wrote: Let $g(x)=f(x)-x^2$, then the functional relation becomes $g(g(x)+x^2+y)=g(g(x)+x^2-y)$ for all real $x,y$. Then $g(g(x)+x^2-g(y)-y^2+z)=g(g(x)+x^2+g(y)+y^2-z)=g(g(y)+y^2-g(x)-x^2+z)$, thus $g(z)=g(2*(g(x)+x^2-g(y)-y^2) +z)$ for all real $x,y,z$. Please help me, I cannot understand. How $z$ came here. There were only $x$ and $y$

stergiu wrote: Find all real functions $f$ defined on $ \mathbb R$, such that \[f(f(x)+y) = f(f(x)-y)+4f(x)y ,\]for all real numbers $x,y$. Note that $f \equiv 0$ is a solution. Let $P(x,y)$ denote the given assertion. Suppose there exists a $t \in \mathbb{R}$, such that $f(t) \neq 0$, then note that $g(x,y) = 2 \cdot f(x) - 2 \cdot f(y)$ is surjective as $P(t, \frac{x}{8 \cdot f(t)})$ yields $x = 2 \cdot f(f(t) + \frac{x}{8 \cdot f(t)}) + 2 \cdot f(f(t) - \frac{x}{8 \cdot f(t)})$. Now note that $P(x, f(x))$ yields that $f(2 \cdot f(x)) = f(0) + 4 \cdot f(x)^2$. Also, $P(x, 2 \cdot f(y) - f(x))$ yields $f(2 \cdot f(x) - 2 \cdot f(y))$ = $f(2 \cdot f(y) - f(x)) - 4 \cdot f(x) \cdot (2f(y) - f(x)) $ = $f(0) + (2 \cdot f(y) - 2 \cdot f(x))^2$. Since $g(x,y)$ is surjective, $f(x) = 2x^2 + c$. It is trivial to verify that all such functions satisfy the given condition.

@above $P(x,0)$ does not give any new information. Can you explain a bit more how do you obtain $f(x)(x^2 - f(x)) = 0$?

Case 1: $f(x)=0.$ Then it's obvious that its a satisfying solution. Case 2: $f(x)\neq 0.$ $P(x,f(x))\implies f(2f(x))=f(0)+4f(x)^2.$ $P(x,2f(fy)-f(x))\implies f(2f(x)-2f(y))=f(2f(y))-4f(x)(2f(y)-f(x))$ $=f(0)+4f(y)^2-4f(x)(2f(y)-f(x))=f(0)+(2f(x)-2f(y))^2.$ $\exists r,z\in \mathbb{R} : f(z)\neq 0$ $\text{and } 2f(f(z)+x/(8f(z)))-2f(f(z)-x/(8f(z)))=r \implies f(r)=r^2+f(0),$ easy to check that it works.

This problem is from Mongolia TST 2020 , hilarious isn't it? Answer: $f(x) = 0$ or $f(x) = x^2 + c$. If for all $x$, $f(x) = 0$ then it's a solution. Thus we can assume there exists $b$ such that $f(b) \neq 0$. Let $P(x, y)$ be the given assertion. Now consider the following claim: Claim: For every $a \in \mathbb{R}$, there exist $u, v \in \mathbb{R}$ such that $f(u) - f(v) = a$. Proof. $P(b, \frac{a}{4f(b)})$ gives $a = f(f(b) + \frac{a}{4f(b)}) - f(f(b) - \frac{a}{4f(b)})$, so we're done. $\blacksquare$ $P(x, f(x))$ yields $f(2f(x)) = f(0) + (2f(x))^2$. Now we'll prove that $f(a) = a^2 + c$ for all $a \in \mathbb{R}$. By claim, there exist $u, v$ such that $\frac{a}{2} = f(u) - f(v)$. Then $P(u, 2f(v) - f(u))$ yields $f(0) + (2f(v))^2 = f(2f(v)) = f(f(u) + 2f(v) - f(u)) = f(2f(v) - 2f(u)) + 4f(u)(2f(v) - f(u))$, or equivalently $f(2f(u) - 2f(v)) = f(0) + (2f(u) - 2f(v))^2$. Hence $f(a) = a^2 + f(0)$ for all $a \in \mathbb{R}$, so we're done. $\blacksquare$

The answer is $f \equiv x^2+c$ or $f \equiv 0$, which all work. Let $A = \{f(x) - f(y) \mid x, y \in \mathbb R\}$. Assume $f \not \equiv 0$ for the rest of the proof. Claim: $A = \mathbb R$. Proof: Fix $x$ such that $f(x) \neq 0$. Then $4f(x)y = f(f(x) + y) - f(f(x) - y) \in A$, and we can vary $y$ to attain any value on the left. $\blacksquare$ Let $c = f(0)$. By setting $y = f(x)$, $f(2f(x)) = f(0 ) + 4f(x)^2$ for all $x \in \mathbb R$. However, \[f(f(y) + 2f(x) - f(y)) = f(f(y) - 2f(x) + f(y)) + 4f(y)(2f(x) - f(y)).\]Rearranging, \begin{align*} f(2f(y) - 2f(x)) &= f(2f(x)) - 4f(y)(2f(x) - f(y)) \\ &= f(0) + 4f(x)^2 - 8 f(x)f(y) + 4f(y)^2 \\ &= f(0) + (2f(y) - 2f(x))^2. \end{align*}So $f \equiv x^2+c$ on all real numbers of the form $f(x) - f(y)$ where $x, y \in \mathbb R$ and thus on $\mathbb R$. Remark: Using quasi-surjectivity is cool.