IFEO P1 vibes? $P(x,y): f(xf(y)+f(x)+y)=xy+f(x)+f(y)$ $P(0,-f(0)): f(-f(0))=0$ If $f(c)=0$, then $P(c,c): c^2=0$. Hence, $f(0)=0$ and $f$ is injective at $0$. $P(x,0): f(f(x))=f(x)$ For any $y\ne 0$, $P(-\frac{y}{f(y)},y): f(f(-\frac{y}{f(y)}))=-\frac{y^2}{f(y)}+f(-\frac{y}{f(y)})+f(y)$. Hence, $f(y)=\frac{y^2}{f(y)}$ or $f(y)^2=y^2$, obviously still true when $y=0$. $P(-1,-1): f(-1)=-1$. $P(-1,y): f(y-f(y)-1)=f(y)-y-1$. Thus, if $f(a)=-a$, then $f(2a-1)=-2a-1$, so either $2a-1=2a+1$(rej) or $2a-1=-2a-1\implies a=0$. Thus, $f(x)=x\forall x \in \mathbb{R}$, which obviously works.

Solved by eser43 Define $P(x,y)$ as the equation below: $$f(xf(y)+f(x)+y)=xy+f(x)+f(y)$$ $P(0,-f(0))$: $$f(0)=f(0)+f(-f(0))$$ so we have:$$f(-f(0))=0$$ $P(-f(0),-f(0))$: $$f(0)^2=0$$ so $$f(0)=0$$ $P(x,0)$: $$f(x)=f(f(x)) (\star)$$ Assume that there is a real $t$ such that $f(t) \neq t$ We want to prove that $f(x)-x$ is surjective. $P(\frac{x}{t-f(t)}+1,t)$: $$f(Z)-Z=x$$ when $Z$ is whats inside the $f(xf(y)+f(x)+y)$ after setting $P(\frac{x}{t-f(t)}+1,t)$. so $f(x)-x$ is indeed surjective. now we have there exists $r_x$ for all $x$ such that $f(r_x)-r_x=x-f(-1)$. $P(r_x,-1)$:$$f(r_xf(-1)+f(r_x)-1)=f(r_x)-r_x + f(-1)=x$$ this proves that $f(x)$ is surjective . So there exists a $t'$ such that $f(t')=t$. In $(\star)$ we put $x\rightarrow t'$: $$f(t)=t$$Contraction. so the assumption about the existence of $t$ is wrong. so we have for all real $x$ :$$f(x)=x$$

My continuation from $f(0)=0:$ $P(-1,y+f(x)+xf(y))\implies f(-xy-f(y)+y+xf(y)+f(-1))=x(y-f(y))-y+f(y)+f(-1)$ Easily, $f(x)\equiv x$ is a solution. Otherwise, there exists $y_0$ such that $f(y_0)\ne y_0$. Then, taking $y=y_0$ and varying $x$ in the above equation, we have $f$ surjective. $P(x,0)\implies f(f(x))=f(x)$. $P(f(x),f(y))\implies f(f(x)f(y)+f(x)+f(y))=f(x)f(y)+f(x)+f(y)$. Using surjectivity, $f(xy+x+y)=xy+x+y$, and since $xy+x+y$ can span through all real numbers $\implies f(x)\equiv x$.(This contradicts $f(y_0)=y_0$) Thus the only solution is $f(x)\equiv x$.

Define $P(x,y)$ as usual. Note that $P(-1,-1):f(-1)=-1$. Let $f(0)=c$.We have: $P(-1,0):f(-c-1)=c-1$ so $P(0,-c-1):-1=2c-1$ and $f(0)=0$. Now by $P(x,0)$ we have $f(f(x))=f(x)$ .On the other hand , if $P(t)=0$ , by $P(t,t)$ we have $t=0$. Now consider a nonzero $y$ : $P(-\frac{y}{f(y)} , y) :f(f(-\frac{y}{f(y)}))=f(y) - \frac{y^2}{f(y)} + f(-\frac{y}{f(y)})$ so multiplying by $f(y)$ , we have $f(y)^2=y^2$. The rest is easy

Sol as everyone .

FALSE SOLUTION

strong_boy wrote: $P(x,A) : f(f(x)+A)= -xA+f(x)$ then $f :$ injective Uhh ? And what if $A=0$ ? (and in fact it is)

Let $P(x,y)$ denote the assertion $f(xf(y)+f(x)+y)=xy+f(x)+f(y)$. By $P(0,-f(0))$, $f(k)=0$ for some $k$. Then $P(k,k)$ gives $k=0.$ So $f(f(x))=f(x)$ and $P(-\tfrac{x}{f(x)},x)$ implies $f(x)\equiv \pm x$. Let $f(v)=-v$ then $P(f(x),v)$ shows $v=0.$

the only answer f(x)=x

We claim that $f(x)=x$ for all $x\in \mathbb{R}$ is the only solution. First, note that, $P(0,-f(0))$ gives \[f(-f(0))=0\]Now, we have the following key claim. Claim : $f((x)=0$ if and only if $x=0$. Proof : Consider $\alpha \in \mathbb{R}$ such that $f(\alpha)=0$. Then, $P(\alpha,\alpha)$ implies \[f(\alpha f(\alpha)+f(\alpha) + \alpha)=\alpha^2 + f(\alpha) +f(\alpha) \implies f(\alpha)=\alpha^2 \implies \alpha =0\]Further, we know that indeed exists $\alpha \in \mathbb{R}$ such that $f(\alpha)=0$ and thus, we must have that $f(0)=0$ and $f(x)=0$ if and only if $x=0$ as desired. Now note that, $P(x,0)$ gives \[f(f(x))=f(x)\]for all $x\in \mathbb{R}$. Now, we look at $P(\frac{-y}{f(y)},)y$ for $y \neq 0$ (note that $\frac{-y}{f(y)} \in \mathbb{R}$ since $f(y) \neq 0$ for all $y\neq 0$ by the above claim). This yields, \begin{align*} f\left(\frac{-y}{f(y)}f(y)+f\left(\frac{-y}{f(y)}\right)+y\right) &= \frac{-y}{f(y)}y + f\left(\frac{-y}{f(y)}\right) + f(y)\\ f\left(f\left(\frac{-y}{f(y)}\right)\right) &= f\left(\frac{-y}{f(y)}\right)+f(y)-\frac{y^2}{f(y)}\\ f\left(\frac{-y}{f(y)}\right) &= f\left(\frac{-y}{f(y)}\right) + f(y) - \frac{y^2}{f(y)}\\ f(y) &= \frac{y^2}{f(y)}\\ f(y)^2 &= y^2 \end{align*}Thus, $f(x)=x$ or $f(x)=-x$ for each $x \in \mathbb{R}$. Now, we consider $x\neq y$ such that $f(x)=x$ and $f(y)=-y$. Then, $P(x,y)$ yields, \[f(-xy+x+y)=xy+x-y\]Now, we know that $f(-xy+x+y)=xy-x-y$ (which implies $x=0$) or $f(-xy+x+y)=-xy+x+y$ (which implies $y=0$ or $x=1$). Thus, $f(x)=x$ for all $x$, $f(x)=-x$ for all $x$, or $f(x)=-x$ for all $x\neq 1$ and $f(1)=1$. Now, we can check that $f(x)=x$ for all $x$ works and $f(x)=-x$ for all $x$ does not. For the remaining case, plugging in $P(2,2)$ yields a contradiction. Thus, the only possible solution is \[f(x) = x \text{ for all } x\in \mathbb{R}\]

My solution is just a bit different than the others. It's redacted this way bc I took it from a project and was too lazy to change it jeje: If $x = y = -1$ (so we can cancel $xf(y) + f(x)$): \[ f(-f(-1) + f(-1) - 1) = 1 + 2f(-1) \implies f(-1) = -1. \]If $y = 0$ (so we can cancel some stuff): \[ f(xf(0) + f(x)) = f(x) + f(0). \]Then, if $x = -1$: \[ f(-f(0) - 1) = f(0) - 1. \]This might look useless at first, but it was very natural to arrive at it, so you might have it on a small corner in your paper. Now, the right hand side of the equation is symmetric, and so \[ f(xf(y) + f(x) + y) = f(yf(x) + f(y) + x). \]If $x=0$ in this equation: \[ f(y + f(0)) = f(yf(0) + f(y)). \]If $y = -1-f(0)$ (so we can cancel $y + f(0)$): \[ f(-1) = f(-f(0) - f(0)^2 + f(-1-f(0))), \]but $f(-1-f(0)) = -1 + f(0)$ (surprise surprise! It wasn't that useless), so $-1 = f(-f(0)^2 - 1)$. We now have another value $t$ such that $f(t) = -1$. If $t \neq -1$, letting $x = -1$ and $y = t$: \[ f(-f(t) + f(-1) + t) = -t + f(-1) + f(t) \implies f(t) = -t-2 \implies t = -1, \]but this is absurd. Therefore $t = -1$ and so $f(0)^2 - 1 = -1$, or $f(0) = 0$. If $y = 0$, $f(f(x)) = f(x)$. Now, how can I cancell $xf(y) + y$ to get only $f(f(x))$? We need $x = \dfrac{-y}{f(y)}$, but there's an issue if $f(y) = 0$. Suppose there exists a number $t \neq 0$ with $f(t) = 0$. If $x = y = t$: \[ f(tf(t) + f(t) + t) = 0 = t^2 + f(t) + f(t) = t^2 \implies t = 0, \]but this is absurd. So we can take $y \neq 0$ and $x = \dfrac{-y}{f(y)}$, from where: \[ f\left(f\left( \dfrac{-y}{f(y)}\right)\right) = \dfrac{-y^2}{f(y)} + f\left(\dfrac{-y}{f(y)}\right) + f(y), \]And as $f(f(x)) = f(x)$, \[ \dfrac{y^2}{f(y)} = f(y) \implies f(y)^2 = y^2 \implies f(y) = \pm y. \]It's easy to check $f(x) = x$ is a solution, but $f(x) = -x$ is not. Can they both happen? If there exists some number $b$ such that $f(b) = -b$, letting $x = y = b$: \[ f(-b^2) = b^2 - 2b. \]Now, $f(-b^2) = -b^2 \, b^2$. In the first case we get $b = 0 \, 1$, and on the second case $b = 0$. But if $b = 1$, $f(1) = -1$, and therefore $1 = -1$, which is absurd. This means $f(x) = x$ for all reals $x$ is the only solution.

$P(0,-f(0)) : f(-f(0))=0 \rightarrow \exists a$ such that $ f(a)=0 $ $P(a,a) : f(a)=a^2 \rightarrow a=0 \rightarrow f(0)=0$ $P(x,0): f(f(x))=f(x)$ and $P(- \frac{y}{f(y)},y) : f(f(- \frac{y}{f(y)})) = - \frac{y^2}{f(y)} + f(- \frac{y}{f(y)} ) + f(y) $ so $ \frac{y^2}{f(y)} = f(y) \rightarrow f(y)^2=y^2 \rightarrow f(y)=\pm y$ $f(x)=-x$ doesn't work in $P(x,y)$ and only possible answer is : $$f(x)=x$$

Nartku wrote:

That's how i solve every problem on FE anyway

Solved with Infinitum, Retrofuel, Apothem, FastTortoise and others from Discord. (Main progress by me and Retrofuel) Claim 1: $f(0) = 0$. Proof: $P(0, -f(0), P(-f(0), -f(0))$. Claim 2: $f(-1) = -1$. Proof: $P(-1, -1)$. Claim 3: $f(f(x)) = f(x)$. Proof: $P(x, 0)$. Claim 3: $f(x)^2 = x^2 \forall x \ne 0$. Proof: $P \left(\frac{-x}{f(x)}, x \right)$. Claim 4: $f(x) = x \forall x$. Proof: Assume otherwise, then $\exists b : f(b) = -b$, then $P(b, b)$ works. Since $f(x) = x$ works, we're done.