Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: introduction for the IMO ShortList/LongList project and regardingsolutions

Making $x=f(y)$, we get $f(f(y))=\frac{f(0)+1}2-\frac{f(y)^2}2$, so if we denote $\frac{f(0)+1}2$ by $\alpha$, we have $f(t)=\alpha-\frac{t^2}2\ (*)$ for all $t\in f(\mathbb R)$. Take some $x=f(z)+f(y)\in f(\mathbb R)+f(\mathbb R)$. Then $f(x-f(y))=\alpha-\frac{(x-f(y))^2}2$ and $f(f(y))=\alpha-\frac{f(y)^2}2$, so if we plug these two into the initial equation we get $f(x)=1-\frac{x^2}2$, which must thus hold for all $x\in f(\mathbb R)+f(\mathbb R)$. Now fix any $x\in f(\mathbb R)+f(\mathbb R)$ in the initial equation. We have $f(x-f(y))=\alpha-\frac{(x-f(y))^2}2$, so $(*)$ holds for all $t\in f(\mathbb R)+f(\mathbb R)-f(\mathbb R)$. Since $-xf(y)+1=f(f(y))+f(x)-f(x-f(y))\in f(\mathbb R)+f(\mathbb R)-f(\mathbb R),\ (*)$ holds for all $t$ of the form $-xf(y)+1$, and by choosing a $y$ with $f(y)\ne 0$ ($f$ is clearly not identically zero) and making $x$ vary over the reals, we find that every real has that form, so $(*)$ actually holds for all $t\in\mathbb R$, and since $f(x)$ takes the value $1-\frac{x^2}2$ when $x\in f(\mathbb R)+f(\mathbb R)$, we find $\alpha=1$, and $f(t)=1-\frac{t^2}2,\ \forall t\in\mathbb R$. Conversely, it's easy to check that this is actually a solution, i.e. it satisfies the functional equation.

When you say $-xf(y)+1 = f(f(y)) + f(x) - f(x-f(y)) \in f(\mathbb R) + f(\mathbb R) - f(\mathbb R)$ where did you get that range? Also, how did you conclude that (*) holds for all $t$ in the form $-xf(y)+1$?

Let $f(z)=0$ for some real number $z$. Then set $y=z$ it will become $f(x-f(z))=f(f(z))+xf(z)+f(x)-1$ (since $f(z)=0$) then we will get $f(0)=1$ $(i)$ Set $x=0$, then $f(0-f(y)=f(f(y)+0\times f(y)+f(0)-1$ using $(i)$ we will get that $f(f(-y))=f(f(y))$ $(ii)$ so our function is even function (this only satisfy our solution ). Now here is: $f(x+f(y))=f(x-(-f(y)))=f(-f(y))-xf(y)+f(x)-1$ $(iii)$ $f(x)=f(x+f(y)-f(y))=f(f(y))+(x+f(y))f(y)+f(x+f(y))-1$ then $f(x)=f(f(y))+(x+f(y))f(y)+f(x+f(y))-1$ now using $(iii)$ we will get that $f(f(y))=1-\frac{f^2(y)}{2}$ As before we have taken $f(z)=0$. If it really exist in our function then ita true. Set $f(y)=z$ then we will have that $z=-\sqrt 2$ or $\sqrt 2$. this two numbers is real numbers and our function is even and bot of them satisfy. Set $f(y)=x$ then $f(x)=1-\frac{x^2}{2}$. so answer will be $f(x)=1-\frac{x^2}{2}$ for all real $x$.

abdurashidjon wrote: $f(f(-y))=f(f(y))$ $(ii)$ so our function is even function I'm not sure if we can say that. abdurashidjon wrote: Set $f(y)=x$ then $f(x)=1-\frac{x^2}{2}$. so answer will be $f(x)=1-\frac{x^2}{2}$ for all real $x$. I think that's only valid if there's a $y$ with $x=f(y)$, and we still didn't know anything about the range of $f$. If you wanted to avoid the assumption that there exists a $z$ with $f(z)=0$, you could have followed the next path: $x=2f(y)$, $f(\frac{x}{2})=f(\frac{x}{2})+\frac{x^2}{2}+f(x)-1$, $f(x)=1-\frac{x^2}{2}$. But of course, this is only valid if there's a $y$ with $x=2f(y)$. So, we still should work a little harder before assuming anything about the range of $f$.

lordWings wrote: abdurashidjon wrote: $f(f(-y))=f(f(y))$ $(ii)$ so our function is even function I'm not sure if we can say that. . sorry for mistake it is like $f(-f(y))=f(f(y))$ this if you check you will get the next i will post later Abdurashid

abdurashidjon wrote: Let $f(z)=0$ for some real number $z$. How can you admite that ??

Truely we can take $f(z)=0$ for any $z$. Because there can be such that $z$. And then we will look for it.

we take $ S=\{f(a)-f(b)|\ (a,b)\in R^{2}\},E=\{f(a)|\ a\in R\}$. $ x=f(y),f(0)=c$ gives $ c=2fof(y)+(f(y))^{2}-1$ $ changing\ x\ by\ f(x)$ gives $ {f(f(x)-f(y))=fof(y)+f(x)f(y)+fof(x)-1=\frac{2(c+1)-(f(x)-f(y))^{2}}{2}-1}$ so $ \forall x\in S: \ f(x)=c-\frac{x^{2}}{2}$ we find easly that $ f$ is not a constante so, $ \exists m\in R: \ f(m)\neq 0$ and $ \forall x\in R: \ xf(m)+fof(m)-1=f(x-f(y))-f(x)\in S$ $ g: R\to R,g(x)=xf(m)+fof(m)-1$ is bijiectiv so, $ R=g(R)\subseteq S$ then $ S=R$ and $ \forall x\in S=R: \ f(x)=c-\frac{x^{2}}{2}$ $ x=y=0$ gives $ f(-c)=f(c)+c-1$ then $ c=1+f(-c)-f(c)=1$ $ \forall x\in S=R: \ f(x)=1-\frac{x^{2}}{2}$

mdk wrote: Determine all functions f : R → R such that $ f (x-f (y)) = f (f (y))+x f (y)+f (x)-1$ for all x, y ε R First we solve the functional equation: $ f(x-t)=f(t)+xt+f(x)-1$ If $ x=t=0\implies f(0)=1$, if $ x=t\implies f(0)=2f(x)+x^{2}-1\implies f(x)=1-\frac{x^{2}}{2}$ If $ x=f(y)\in Im f$ then $ c=f(0)=f(x)+x^{2}+f(x)-1\implies f(x)=-\frac{x^{2}}{2}+\frac{c+1}{2}$, where $ c=f(0)$. So $ t=f(y)\in Im f$ therefore $ f(x-t)=-\frac{t^{2}}{2}+\frac{c-1}{2}+xt+f(x)$ If $ x-t=y\in Imf\implies-\frac{y^{2}}{2}+\frac{c+1}{2}=f(y)=-\frac{t^{2}}{2}+\frac{c-1}{2}+yt+t^{2}+f(y+t)\implies f(y+t)=-\frac{(y+t)^{2}}{2}+1$ so $ f(x)=-\frac{x^{2}}{2}+1, x\in Imf+Imf\implies Imf+ImF\subset(-\infty,1]\implies \forall x\in \mathbb{R}\; \exists y,t\in Im f, x=y+t\implies f(x)=-\frac{x^{2}}{2}+1, \forall x\in \mathbb{R}$

I found a solution to this problem in 2 pages. It requires a really interesting trick. I am hoping that anyone can come up with a simpler solution. For those who are interested in my solution you can find it at: http://www.4shared.com/document/krpZ5Oeg/IMO_1999_-_6_Solution.html Link to the original problem: http://www.artofproblemsolving.com/Forum/resources.php?c=1&cid=16&year=1999&sid=5777a41e27950f68ccb7f28b8760fa1b

Is following solution true? Let $f(o)=c$, $f(y)=a$ put $x=a$, then \[f(a)=\frac{c+1-a^2}{2}=f(-a)\] put $x=0$,then \[f(-a)=f(a)+c-1\] then $c=1$ and \[f(x)=1-\frac{x^2}{2}\]

NO. Not to mention that the fact that $f(-a) = \frac {c+1-a^2} {2}$ is not justified, the final conclusion would in fact be $f(a) = 1 - \dfrac {a^2} {2}$, i.e. $f(f(y)) = 1 - \dfrac {f(y)^2} {2}$, or, if you want to change variables, $f(f(x)) = 1 - \dfrac {f(x)^2} {2}$. This means the quadratic form you propose as unique solution holds over $\textrm{Im} f = f(\mathbb{R})$. How do you propose to show it holds over the whole $\mathbb{R}$ ? If you just looked at the first proof provided (post #2), you would have noted that this is the crucial issue ... so your attempt is just a simple first step towards a much more sofisticated full solution.

Thanks Mavropnevma. I understand that it was true only for x-es that are in the set of values of the given function(is there a name of this set?) PS: it was not my solution. I can't find even incomplete solution like this. PS: What is the difference between range of a function and set of values of a function? Or they are equal sets? Sorry, i am beginner.

I don't quite think that you can admit that there exists a real $z$ such that $f(z)=0$ (eg: $f(x)=x^2+1$ doesn't have zeroes on reals, the same way you didn't know if the function in the problem had zeroes), but this problem can be solved preety easilly knowing a quite simple and usefull trick. First of all simple check can show that the function can't be zero on the whole domain so take $t$ such that $f(t)=a$ is not $0$. Putting $y=t$ we get: $f(x-a)-f(x)=ax+f(a)-1$, so we have that every real number $x$ can be written as $x=f(u)-f(v)$, for some real numbers $u,v$. Putting $y=f(y)$ we get: $f(f(x)-f(y))=f(f(x))+f(x)f(y)+f(f(y))-1$. Putting $x=y$ in this relation we get: $f(f(x))=\cfrac{1}{2}\cdot (1+f(1)-f(x)^2)$. Introducing this relation in the previous one we get that: $f(f(x)-f(y))=f(0)-\cfrac{1}{2}\cdot (f(x)-f(y))^2$. So by putting $x=u,y=v$ we get that: $f(x)=f(0)-\cfrac{x^2}{2}$. A simple verification leads us to the single solution: \[f(x)=1-\frac{x^2}{2}\]

Let $\Omega = \text{Img}(f)$ and $g(x)=f(x)-(1-\frac{x^2}{2})$. By letting $x \in \Omega$ we verify that $f(0)=1$ and by letting $x=f(y)$ we verify $g(\omega)=0 \forall \omega \in \Omega$. Notice that $g(x)=g(x+\omega) \forall x\in \mathbb{R},\omega \in \Omega$, therefore $g$ is periodic with period $\omega_1-\omega_2$ for all $\omega_1,\omega_2 \in \Omega$. Finally notice that if $\omega_1 = f(x_1),\omega_2=f(x_2)$ with $x_2=x_1+\omega$ for any $\omega \in \Omega,\omega \neq 0$ (for example, $\omega=1$) then $\omega_1 - \omega_2 = x \omega - (\omega^2/2)$, which can take any real value. This implies $g$ is periodic modulo anything, so it is constant. Since it takes the value $0$, $g \equiv 0$ and so $f \equiv 1-(x^2/2)$

Denote $F=\{f(y)| y \in \mathbb{R}\}$ and $A=\{a-b|a,b\in F\}$. First, for $b \in F$, plugging $x=a$ and $f(y)=b$ to the equation gives $f(a-b)=f(b)+ab+f(a)-1$ for all $a \in \mathbb{R}$ and $b \in F$. Setting $a=b$ gives $f(0)=2f(a)+a^2-1$, so $f(a)=-\frac{1}{2}a^2+\frac{f(0)+1}{2}$ for all $a \in F$. For all $c$ such that $f(c) \not= 0$, we have $f(x-f(c))-f(x)=f(f(c))+xf(c)-1$. Since the R.H.S can take all real values, we have $A \equiv \mathbb{R}$. Now for $a, b \in F$, we have $f(a-b)=ab+f(a)+f(b)-1=ab-\frac{1}{2}a^2-\frac{1}{2}b^2+f(0)+1-1=-\frac{1}{2}(a-b)^2+f(0)$. Therefore, $f(x)=-\frac{1}{2}x^2+f(0)$ for all $x \in \mathbb{R}$. Note that $\frac{f(0)+1}{2}=f(0)$, so $f(0)=1$. This gives $f(x)=-\frac{1}{2}x^2+1$ as our only answer.

Let $x = f(y).$ Then \[f(0) = f(f(y)) + f(y)^2 + f(f(y)) - 1 = 2f(f(y)) + f(y)^2 - 1.\] Let $t = f(y).$ Then \[f(0) = 2f(t) + t^2 - 1.\] Thus, \[f(t) = \dfrac{f(0) + 1}{2} - \dfrac{t^2}{2}\] Replace $t$ with $x - f(y).$ Then \[f(x - f(y)) = \dfrac{f(0) + 1}{2} - \dfrac{x^2 - 2xf(y) + f(y)^2}{2} = f(f(y)) + xf(y) + f(x) - 1.\] Thus, \[\dfrac{f(0) + 1}{2} = \dfrac{x^2}{2} - xf(y) + \dfrac{f(y)^2}{2} + f(f(y)) + xf(y) + f(x) - 1.\] This simplifies down to \[\dfrac{f(0) + 1}{2} = \dfrac{x^2}{2} + \dfrac{f(y)^2}{2} + f(f(y)) + f(x) - 1.\quad\quad(1)\] Replace $t$ with $f(y).$ Then \[f(f(y)) = \dfrac{f(0) + 1}{2} - \dfrac{f(y)^2}{2}.\] Substituting this into $(1),$ we get \[\dfrac{f(0) + 1}{2} = \dfrac{x^2}{2} + \dfrac{f(y)^2}{2} + \dfrac{f(0)+1}{2} - \dfrac{f(y)^2}{2} + f(x) - 1.\] There are many terms that cancel, leading us to the equation \[\dfrac{x^2}{2} + f(x) - 1 = 0.\] This is \[f(x) = 1 - \dfrac{x^2}{2}.\] Replacing this back into the original equation, we get \begin{align*} 1 - \dfrac{x^2 - 2xf(y) + f(y)^2}{2} &= 1 - \dfrac{1 - y^2 + \frac{y^4}{4}}{2} + x - \dfrac{xy^2}{2} + 1 - \dfrac{x^2}{2} - 1\\ 1 - \dfrac{x^2}{2} + x - \dfrac{xy^2}{2} - \dfrac{1-y^2+\frac{y^2}{4}}{2} &= - \dfrac{1-y^2+\frac{y^2}{4}}{2} + x - \dfrac{xy^2}{2} + 1 - \dfrac{x^2}{2}. \end{align*} Everything cancels out, so $\boxed{f(x) = 1 - \dfrac{x^2}{2}}$ for all $x\in \mathbb{R}.$

sturdyoak2012 wrote: Replace $t$ with $x - f(y).$ I don't quite think that you can do this, as you defined $t$ to be a member of the image of $f$. For example, if $f\equiv0$, then $t$ can only be $0$, so then when we make this substitution, $x$ must be $0$ (then $x$ cannot range all real numbers).

Guided Solve:

$f(x-f(y)) = f(f(y)) + xf(y) + f(x) - 1$ (equation 1) $f(x) = 1 - \dfrac {x^2} 2$ satisfies the relation, so that's one possible answer. If $f(y)=0$ for some $y$, then from equation (1), $f(0)=1$. So $f(0)\neq 0$. Then setting $x=\dfrac 1 {f(0)}, y = 0$ in equation 1, $f\left(\dfrac 1 {f(0)} - f(0)\right) = f(f(0)) + f\left(\dfrac 1 {f(0)}\right)$ (equation 2) For $r\in \text{range(f)}$, setting $x=r, f(y)=r$ in equation 1, we get $f(r) = \dfrac {1+f(0) - r^2} 2$ For $r_1,r_2 \in \text{range}(f)$, setting $x=r_1+r_2, y=r_2$ in equation 1, we get $f(r_1+r_2) = 1 - \dfrac {(r_1+r_2)^2} 2$. But the left side of equation 2 is of the form $f(x)$, and the right side is of the form $f(x_1)+f(x_2)$. So applying $f$ to both sides of equation 2, we get that $f(0)=1$. Let $S$ be the set of $x$ such that $f(x) = 1-\dfrac {x^2} 2$. Then for $s\in S$ and $r\in \text{range}(f), s-r\in S$ and $s+r\in S$, as can be verified from equation 1. $f(x-1) = f(x-f(0)) = f(1) + x + f(x) - 1$, so $x = f(x-1) - f(1) - f(x) + f(0)$. So $x\in S$. So $f(x) = 1 - \dfrac {x^2} 2$.

we can solve it directly by putting $x=2f(y)$ is this right??

Here is my solution for this problem Solution $f(x - f(y)) = f(f(y)) + xf(y) + f(x) - 1$ It's easy to see that: $f(x) \equiv 0$ doesn't satisfies the problem Suppose there exist $y_0 \in \mathbb{R}$ which satisfies $f(y_0) \ne 0$ Let $y = y_0$, we have: $f(x - f(y_0)) = f(f(y_0)) + xf(y_0) + f(x) - 1$ or $f(x - f(y_0)) - f(x) = f(f(y_0)) + xf(y_0) - 1$ Then with $x \in \mathbb{R}$, there exist $u, v \in \mathbb{R}$ which satisfy $x = f(u) - f(v)$ Let $x$ be $f(y)$, we have: $f(f(y)) = \dfrac{- f^2(y) + f(0) + 1}{2}$ Let $x$ be $f(u)$, $y$ be $f(v)$, we have: $f(x) = f(f(u) - f(v)) = f(f(v)) + f(u)f(v) + f(f(u)) - 1 = - \dfrac{f^2(u)}{2} + f(u)f(v) - \dfrac{f^2(v)}{2} + f(0) = - \dfrac{1}{2} (f(u) - f(v))^2 + f(0) = - \dfrac{x^2}{2} + f(0)$ So: $f(x) = - \dfrac{x^2}{2} + a, \forall x \in \mathbb{R}, a \in \mathbb{R}$ Then: $\dfrac{- f^2(x) + a + 1}{2} = f(f(x)) = - \dfrac{f(x)^2}{2} + a$ or $a = 1$ Hence: $f(x) = - \dfrac{x^2}{2} + 1, \forall x \in \mathbb{R}$ Retry, we see that: $f(x) = - \dfrac{x^2}{2} + 1$ satisfies the problem In conclusion, we have: $f(x) = - \dfrac{x^2}{2} + 1, \forall x \in \mathbb{R}$

I think letting g(x)=f(x)-1 can be good but could not get the answer can someone continue this ?

Its easy to check that $f \equiv 0$ is not a solution $\implies \exists x_0\in R$ such that $f(x_0)\neq 0 $ Now we put $\frac{x-f(f(x_0))+1}{f(x_0)}$ and $x_0$ in the original equation: $f(\frac{x-f(f(x_0))+1}{f(x_0)}-f(x_0))-f(\frac{x-f(f(x_0))+1}{f(x_0)})=x\implies f(x)-f(y)$ runs through all real values $(*)$. Let $g(x)=f(x)+\frac{x^2}{2}-1$, after replacing $f$ to $g$ in the original equation we get $g(x-g(y)+\frac{y^2}{2}-1)=g(g(y)-\frac{y^2}{2}+1)+g(x) (**)$ and because of $(*)$ we get that $g(x)-\frac{x^2}{2}-g(y)+\frac{y^2}{2}$ also runs through all real values $(***)$ Now we place $x=g(y)-\frac{y^2}{2}+1$ to the $(**)$: $g(g(y)-\frac{y^2}{2}+1)=\frac{g(0)}{2}$ Then we place $x=g(x)-\frac{x^2}{2}+1$ to the $(**)$: $g(g(x)-\frac{x^2}{2}-g(y)+\frac{y^2}{2})=g(g(x)-\frac{x^2}{2}+1)+g(g(y)-\frac{y^2}{2}+1=g(0)$, in respect that $(***)$ we conclude that $g$ is constant $\implies f(x)+\frac{x^2}{2}-1+c=0$, after substitution to the original equation we get $c=0$ So there is only one solution: $f(x)=-\frac{x^2}{2}+1$

mavropnevma wrote: NO. Not to mention that the fact that $f(-a) = \frac {c+1-a^2} {2}$ is not justified, the final conclusion would in fact be $f(a) = 1 - \dfrac {a^2} {2}$, i.e. $f(f(y)) = 1 - \dfrac {f(y)^2} {2}$, or, if you want to change variables, $f(f(x)) = 1 - \dfrac {f(x)^2} {2}$. This means the quadratic form you propose as unique solution holds over $\textrm{Im} f = f(\mathbb{R})$. How do you propose to show it holds over the whole $\mathbb{R}$ ? If you just looked at the first proof provided (post #2), you would have noted that this is the crucial issue ... so your attempt is just a simple first step towards a much more sofisticated full solution. Can you please tell me whether my solution is correct or wrong if wrong where I made mistake? $f(x-f(y))=f(f(y)) +xf(y)+f(x)-1$ .....1 If $f(0)=0$ we put $y=0$ then 1 becomes $f(x)=f(x)-1$, a contradiction and thus $f(0)=c$, a non-zero real number And $f(x-c)=f(c)+cx+f(x)-1$ for all x in R So we have $x=\dfrac{f(x-c)-f(x)-f(c)+1}{c}$......2 and x can't be constant since it is true for all x in R So for any x in R(codomain) we have pre-image of f, and thus f is surjective. Thus we can have x_0 in R such that $f(x_0)=0$ and put it in original equation 1, $f(x-f(x_0)=f(f(x_0))+xf(x_0)+f(x)-1$ $f(x)=f(0)+f(x)-1$ implies $f(0)=1$ Now put $x=f(y)$ for any x in R (surjective) in original equation 1 then $f(0)=2f(x)+(x^2)-1$ and we get, $f(x)=1-\dfrac{x^2}{2}$ If anyone find out any mistake plz tell me it will be helpful, Thank you

Ashim wrote: ... and thus f is surjective. ... ... and we get $f(x)=1-\dfrac{x^2}{2}$ Aren't you a little bit worried by getting a non surjective final result after having "proved" that $f(x)$ was surjective (and after having used this - wrong - assertion to get your non surjective result) ?

Let $P(x,y)$ be the assertion $f(x-f(y))=f(f(y))+xf(y)+f(x)-1$. If $f(0)=0$ then $P(0,0)\Rightarrow0=-1$, contradiction. So $f(0)\ne0$. $P(x,0)\Rightarrow f(x-f(0))-f(x)=xf(0)+f(f(0))-1$ and since $f(0)\ne0$, the RHS is surjective. Thus, $\forall x\exists u,v:f(u)-f(v)=x$. $P(f(u),v)\Rightarrow f(f(u)-f(v))=f(f(v))+f(u)f(v)+f(f(u))-1$ $P(f(x),x)\Rightarrow f(0)=2f(f(x))+f(x)^2-1\Rightarrow f(f(x))=\frac12\left(f(0)+1-f(x)^2\right)$ So: \begin{align*} f(x)&=f\left(f(u)-f(v)\right)\\ &\overset{P(f(u),v)}{=\joinrel=\joinrel=\joinrel=\joinrel=}\frac12\left(f(0)+1-f(v)^2\right)+\frac12\left(f(0)+1-f(u)^2\right)+f(u)f(v)-1\\ &=f(0)-\frac12\left(f(u)-f(v)\right)^2\\ &=f(0)-\frac12x^2 \end{align*}Testing, we see that $\boxed{f(x)=1-\frac12x^2}$ is the only solution..

Firstly we can get the function on simply putting values which is $f(x)=1-\frac{x^2}{2}$ , It is not surjective nor Injective Now to show that the above $f(x)$ is the only solution By putting $x=y=0$ $$f(-f(0))=f(f(0))+f(0)-1$$ We can see that $f(0) \not = 0$ Lets take $f(y)=a$ and $x=b$ where $a$ is a $R$ belonging to the range and $b \in R$ Lets take $a=b$ then we get $f(0) = f(a)+a^2+f(a)-1$ , \begin{align} f(a)= \frac{1-a^2+k}{2} \end{align} where $f(0)=k$ Using $f(y)=k$ and $x=b$ in the original equation , we get $f(b-k)=f(k)+bk+f(b)-1$ $$ f(b-k)-f(k)=bk+f(b)-1 $$ as $b \in R$ take $b=\frac{x+1-f(k)}{k}$ we get $$ f( \frac{x+1-f(k)}{k} - k) - f(k) = x $$ So we can write $x$ as a differenece of two numbers $c,d$ where $c,d$ belong in the range take $x=c$ , $f(y)=d$ , we get $$ f(x) = f(d)+cd+ f(c)-1$$ Using (1) on $f(c)$ and $f(d)$ $$f(x)= \frac{1-d^2+k}{2} +\frac{2cd}{2} + \frac{1-c^2+k}{2} -\frac{2}{2}$$$$f(x) = \frac{2k-c^2-d^2+2cd}{2}$$$$f(x) = k - \frac{(c-d)^2}{2}$$ $$f(x)= k -\frac{x^2}{2}$$(2) now just take $f(a)$ and Use (1) =(2) we get $k=1$ and $$f(x)=1-\frac{x^2}{2}$$

The answer is $f(x) = -\frac{1}{2} x^2+1$ which obviously works. For the other direction, first note that \[ P(f(y),y) \implies 2f(f(y)) + f(y)^2 - 1 = f(0). \]We introduce the notation $c = \frac{f(0)-1}{2}$, and $S = \operatorname{Image} f$. Then the above assertion says \[ f(s) = -\frac{1}{2} s^2 + (c + 1). \]In light of this, the given functional equation can be rewritten as \[ Q(x,s) : f(x-s)=-\frac{1}{2} s^2 + sx + f(x) - c. \] Claim: [Main claim] We can find a function $g \colon {\mathbb R} \to {\mathbb R}$ such that \[ f(x-z) = zx + f(x) + g(z). \qquad (\spadesuit). \]Proof. If $z \neq 0$, the idea is to fix a nonzero value $s_0 \in S$ (it exists) and then choose $x_0$ such that $- \frac{1}{2} s_0^2 + s_0 x_0 - c = z$. Then, $Q(x_0, s)$ gives an pair $(u,v)$ with $u-v = z$. But now for any $x$, using $Q(x+v,u)$ and $Q(x,-v)$ gives \begin{align*} f(x-z)-f(x) &= f(x-u+v)-f(x) = f(x+v)-f(x) + u(x+v) - \frac{1}{2} u^2 + c \\ &= -vx-\frac{1}{2} s^2-c + u(x+v) - \frac{1}{2} u^2 + c \\ &= -vx-\frac{1}{2} v^2 + u(x+v) - \frac{1}{2} u^2 = zx + g(z) \end{align*}where $g(z) = -\frac{1}{2}(u^2+v^2)$ depends only on $z$. $\blacksquare$ Now, let \[ h(x) \overset{\text{def}}{=} \frac{1}{2} x^2 + f(x) - (2c+1), \]so $h(0) = 0$. Claim: The function $h$ is additive. Proof. We just need to rewrite $(\spadesuit)$. Letting $x=z$ in $(\spadesuit)$, we find that actually $g(x)=f(0)-x^2-f(x)$. Using the definition of $h$ now gives \[ h(x-z) = h(x) + h(z). \]$\blacksquare$ To finish, we need to remember that $f$, hence $h$, is known on the image \[ S = \left\{ f(x) \mid x \in {\mathbb R} \right\} = \left\{ h(x) - \frac{1}{2} x^2 + (2c+1) \mid x \in {\mathbb R} \right\}. \]Thus, we derive \[ h\left( h(x)-\frac{1}{2} x^2+(2c+1) \right) = -c \qquad \forall x \in {\mathbb R}. \qquad(\heartsuit) \]We can take the following two instances of $\heartsuit$: \begin{align*} h\left( h(2x)-2x^2+(2c+1) \right) &= -c \\ h\left( 2h(x)-x^2+2(2c+1) \right) &= -2c. \end{align*}Now subtracting these and using $2h(x)=h(2x)$ gives \[ c = h\left( -x^2 - (2c+1) \right). \]Together with $h$ additive, this implies readily $h$ is constant. That means $c=0$ and the problem is solved.

Easy for IMO P6? $P(f(x),x): f(0)=2f(f(x))+f(x)^2-1\implies -2f(f(x))=f(x)^2-f(0)-1\implies f(f(x))=-\frac{1}{2}(f(x)^2-f(0)-1)$. $P(0,0): f(-f(0))=f(f(0))+f(0)-1$. If $f(0)=0$, then $0=-1$, so $f(0)\ne0$. $P(x,0): f(x-f(0))-f(x)=f(f(0))+xf(0)-1$, which implies $f(x)-f(x+f(0))$ is surjective as $f(0)\ne0$. So we can find $a$ and $b$ so that $f(a)-f(b)=x$. $P(f(a),b): f(f(a)-f(b))=f(x)=f(f(a))+f(a)f(b)+f(f(b))-1$ We have \[f(x)=-\frac{1}{2}(f(a)^2-f(0)-1)-\frac{1}{2}(f(b)^2-f(0)-1)+f(a)f(b)-1=-\frac{1}{2}f(a)^2+\frac{1}{2}f(0)+\frac{1}{2}-\frac{1}{2}f(b)^2+\frac{1}{2}f(0)+\frac{1}{2}+f(a)f(b)-1=-\frac{1}{2}(f(a)-f(b))^2+f(0)\] So $f(x)=-\frac{1}{2}x^2+f(0)$. $P(0,0): f(-f(0))=f(f(0))+f(0)-1$. Since $f$ is even, we have $f(0)-1=0\implies f(0)=1$. Thus, our answer is $\boxed{f(x)=-\frac{1}{2}x^2+1}$.

Let $P(x,y)$ denote the assertion. We have $(1)$ $P(0,0)$: $f(-f(0))=f(f(0))+f(0)-1.$ $(2)$ $P(x,0)$: $f(x-f(0))=f(f(0))+xf(0)+f(x)-1.$ If we move the terms with $x$ on inside, then we get $f(x-f(0))-f(x)=f(f(0))+xf(0)-1.$ $(3)$ $P(0,y)$: $f(-f(y))=f(f(y))+f(0)-1.$ $(4)$ $P(f(y),y)$: $f(0)=2f(f(y))+f(y)^2-1.$ Rearranging this gives $f(f(y))=c-\tfrac12 f(y)^2$ where $c=\tfrac{f(0)+1}{2}.$ Thus, $f(y)=c-\tfrac12 y^2$ for all $y$ in the range of $f$. $(5)$ $P(f(x),y)$: $f(f(x)-f(y))=f(f(y))+f(x)f(y)+f(f(x))-1.$ Since the entire RHS of this is symmetric, $f(f(x)-f(y))=f(f(y)-f(x)).$ Notice that we have something of the form $f(x)-f(y)$ in $(2)$, so in $(5)$ let $x\to y-f(0)$ to get that if $z=f(f(0))+xf(0)-1$ then $f(z)=f(-z)$. If $f(0)=0$ then $(1)$ is violated. Therefore, $z$ can be any number, so $f$ is even. Then, from $(3)$ we get that $f(0)=1$ so from $(4)$, $f(y)=1-\tfrac12 y^2$ for all $y$ in the range of $f$. From $(5)$, we can get \[f(f(x)-f(y))=1-\frac12 f(y)^2-\frac12 f(x)^2 +f(x)f(y)=1-\frac12 (f(x)-f(y))^2.\]Since from $(2)$, the expression $f(x)-f(y)$ is surjective, $f(x)=1-\tfrac12 x^2$, which is a valid solution.

The answer is $f(x)=1-\frac{x^2}{2}$ only, which clearly works. Let $P(x,y)$ denote the assertion. Obviously $f \equiv 0$ doesn't work. Thus fix $y$ with $f(y) \neq 0$ and vary $x$, rearranging the equation to be $f(x-f(y))-f(x)=xf(y)+(f(f(y))-1)$. This implies that $f(a)-f(b)$ can take on any value in $\mathbb{R}$ as $a,b$ vary in $\mathbb{R}$. On the other hand, by $P(f(x),y)$ we obtain $f(f(x)-f(y))=f(f(y))+f(x)f(y)+f(f(x))-1$, which is symmetric in $x,y$; thus, $f(f(x)-f(y))=f(f(y)-f(x))$, so combining these implies $f$ is even. Now, $P(0,y)$ implies $f(-f(y))=f(f(y))+f(0)-1$, which implies $f(0)=1$ since $f$ is even. Now let $$S=\left\{a \mid a \in \mathbb{R} \text{ and } f(a)=1-\frac{a^2}{2}\right\};$$our goal is to show $S=\mathbb{R}$. From $P(f(y),y)$, it follows that $f(y) \in S$ for all $y$. Now, looking at the original assertion, if $x \in S$ then $$f(f(y))+xf(y)+f(x)-1=1-\frac{f(y)^2}{2}+xf(y)+1-\frac{x^2}{2}-1=1-\frac{(x-f(y))^2}{2},$$so $x-f(y) \in S$. Furthermore, since $f$ is odd, we have $x \in S \implies -x \in S \implies -x-f(y) \in S \implies x+f(y) \in S$ as well. Hence (finite) integer linear combinations of outputs of $f$ are in $S$. Since $f(0)=1$, we have $1 \in S \implies f(1)=\frac{1}{2}$. Thus, from $P(x,0)$ we obtain $$f(x-1)=\frac{1}{2}+x+f(x)-1 \implies f(x-1)-f(x)+\frac{1}{2}=f(x-1)-f(x)+f(1)=x.$$Thus every $x \in \mathbb{R}$ can be written as a linear combination of outputs of $f$, hence all real numbers are in $S$ as desired. $\blacksquare$ Remark: Totally forgot you can just look at $P(f(x),y)$ and evaluate the RHS directly whoops. But I think this solution is also not that bad. $x \in S \implies x-f(y) \in S$ comes from the idea that if we "know" $f(x)$ well, we also "know" $f(x-f(y))$ well. This works because nothing on the LHS is really "wrapped", since we know $f(f(y))$ "unwraps" itself. This is enough to imply the linear combinations fact, but in general this won't finish since it could be the case that, say, $S=\mathbb{Q}$. Logically, though, this shouldn't happen, because if we can only get to rationals then plugging in $x=\pi$ would be a contradiction, and it turns out that this idea leads to the finish.

wait this is kinda cool Let $P(x,y)$ be the assertion. From $P(f(x),y)$ and $P(f(y),x)$ we find \[f(f(x)-f(y))=f(f(x))+f(f(y))+f(x)f(y)-1=f(f(y)-f(x)).\] Now I claim $g(x,y):=f(x)-f(y)$ is surjective; it suffices to show that \[xf(y)+f(f(y))-1=f(x-f(y))-f(x)\]is surjective. This is simple: evidently $f\equiv 0$ is impossible hence choose $f(y)\neq 0$ and vary $x$. Hence we find $f(a)=f(-a)$. Suppose there exists $x$ and $y$ such that $f(y)=2x$; that is, $2x$ is in the range of $f$. Then our previous result and $P(x,y)$ yield \[f(x)=f(-x)=f(2x)+2x^2+f(x)-1\implies f(2x)=-2x^2+1.\]As a result we find \[f(f(x))=-\frac{f(x)^2}{2}+1.\] Returning to $P(f(x),y)$ we obtain \begin{align*} f(f(x)-f(y)) &= \left(-\frac{f(x)^2}{2}+1\right) + \left(-\frac{f(y)^2}{2}+1\right)+f(x)f(y)-1 \\ &= -\left(\frac{f(x)^2+f(y)^2-2f(x)f(y)}{2}\right)+1 \\ &= -\left(\frac{(f(x)-f(y))^2}{2}\right)+1 \end{align*}and since $g(x,y)$ is surjective this simply becomes $f(x)=-\frac{x^2}{2}+1$. A tedious (not really) computation confirms this solution. $\blacksquare$