interesting

Assassino9931 wrote: Determine all functions $f:\mathbb{R} \to \mathbb{R}$ such that \[ f(xy+f(x)) + f(y) = xf(y) + f(x+y) \]for all real numbers $x$ and $y$. Let $P(x,y)$ be the assertion $f(xy+f(x))+f(y)=xf(y)+f(x+y)$ $P(0,0)$ $\implies$ $f(f(0))=0$ $P(f(0),0)$ $\implies$ $2f(0)=f(0)^2$ and so $f(0)\in\{0,2\}$ 1) If $f(0)=2$ (and so $f(2)=0$) $P(x,0)$ $\implies$ $f(f(x))+2=2x+f(x)$ and so $f(x)$ is injective. If $x\ne 1$, $P(x,\frac{f(x)-x}{1-x})$ $\implies$ $f(\frac{f(x)-x}{1-x})=0=f(2)$ and so, since injective, $\frac{f(x)-x}{1-x}=2$ And so $f(x)=2-x$ $\forall x\ne 1$ $P(1,x)$ $\implies$ $f(x+f(1))=f(x+1)$ and so, since injective $f(1)=1=2-1$ And so $\boxed{\text{S1 : }f(x)=2-x\quad\forall x}$, which indeed fits 2) If $f(0)=0$ If $f(a)=f(b)=c$, subtracting $P(a,b)$ from $P(b,a)$, we get $c(a-b)=0$ and so $f(a)=f(b)=0$ or $a=b$ $P(x,0)$ $\implies$ $f(f(x))=f(x)$ and so $f(x)=0$ or $f(x)=x$ (using previous line) If $f(1)=0$, $P(1,x)$ $\implies$ $f(x)=f(x+1)$ and so $\boxed{\text{S2 : }f(x)=0\quad\forall x}$, which indeed fits If $f(1)=1$ : If $\exists x\ne 0$ such that $f(x)=0$, then $P(1-x,x)$ $\implies$ $f(x-x^2+f(1-x))=1$ and so (since nonzero), $f(x-x^2+f(1-x))=x-x^2+f(1-x)$ $\implies$ $x-x^2+f(1-x)=1$ and then : Either $f(1-x)=0$ and this becomes $x-x^2=1$, impossible. Either $f(1-x)=1-x$ and this becomes $x-x^2+1-x=1$, impossible (since $x\ne 0$) So no such $ x$ and $\boxed{\text{S3 : }f(x)=x\quad\forall x}$, which indeed fits

Here is a slightly different solution, hopefully correct.

A bit harder than expected. Put $x=1$ to obtain $f(y+1)=f(y+f(1))$, and so if $f(1) \neq 1$ then $f$ is periodic with period $T=f(1)-1 \neq 0,$ which by putting $y \rightarrow y+T$ results in $f(xy+f(x))=f(xy+f(x)+xT),$ i.e. $f(y)=f(y+xT)$ for all $x,y$, hence $f$ is constant. The only constant function that works is the zero function. Now, assume that $f(1)=1$. Put $y=0$ to obtain $f(f(x))=xf(0)-f(0)+f(x),$ and so $f$ is injective provided that $f(0) \neq 0$. In that case, we put $y=\dfrac{x-f(x)}{x-1}$ for all $x \neq 1$ to obtain that $f(\dfrac{x-f(x)}{x-1})=0,$ and so $\dfrac{x-f(x)}{x-1}=c$ for some constant $c$. Some easy algebra reveals that $c=2$, and so $f(x)=2-x$ for all $x$ in this case. On the other hand, assume that $f(0)=0$. Then, put $y=0$ to obtain $f(f(x))=f(x)$ for all $x$. Moreover, put $x=-1$ and $y=f(-1)$ to obtain $f(-1)=-f(-1)+f(f(-1)-1),$ and so $f(f(-1)-1)=2f(-1)$. Now, put $x=f(-1)$ and $y=-1$ to obtain $f(-1)+f(-1)^2=f(f(-1)-1)$. Equating these two, we obtain $f(-1) \in \{0,-1 \}$. However, if $f(-1)=0$, then $x=-1, y=1$ yields $0+1=-1+0,$ a contradiction. Hence, $f(-1)=-1$, and so $x=-1$ implies $f(-y-1)+2f(y)=f(y-1)$ Putting $y \rightarrow -y$ here gives us $f(y-1)+2f(-y)=f(-y-1)=f(y-1)-2f(y),$ and so $f$ is odd. By putting $x \rightarrow -x$ in the given equation and adding the two sides gives us $f(y+x)+f(y-x)=2f(y),$ which is well known to imply that $f$ is additive. Turning back to the original equation, we conclude that $f(xy)=xf(y),$ and so we may put $y=1$ here to obtain that $f$ is the identity function. To sum up, we have three solutions, the zero, the identity, and $2- \rm id$.

Assassino9931 wrote: Determine all functions $f:\mathbb{R} \to \mathbb{R}$ such that \[ f(xy+f(x)) + f(y) = xf(y) + f(x+y) \]for all real numbers $x$ and $y$.

If $f(xy+f(x))=(x-1)f(y)+f(x+y)$ and $f(xy+f(y))=(x-1)f(x)+f(x+y)$ lead to the conclusion that if $f(x) = f(y) \neq 0$, then $x = y$(*). Substitute $x=1$: $f(y+f(1))=f(y+1)$, which implies $f(x) \equiv 0$(satisfies requirement) or $f(1)=1$. Here we consider $f(1)=1$. Substitute $x=0$: $f(f(0))=0$, hence there exists $a$, such that $f(a)=0$. For any $a$ where $f(a)=0$, substitute $y=a$: $f(ax+f(x))=f(x+a)$. Substitute $x=a$: $f(ay)=(a-1)f(y)+f(a+y)$, then substitute $y=0$, $f(0)=(a-1)f(0)$. So $f(0)=0$ or $a=2$ is the only solution making $f(a)=0$. Case 1: $f(0)=0$. Substitute $y=0$ into the original equation: $f(f(x))=f(x)$, hence for any $x$, $f(x)=x$ or 0. Substitute $y=1$: $f(x+f(x))=x-1+f(x+1)$. Here, substitute $x$ with any $a$ such that $f(a)=0$, $0=a-1+f(a+1)=a-1$ or $2a$. And since $a\neq1$, then $a=0$. Thus, $f(x)=x$ holds for all $x$. Case 2: $a=2$ is the only solution making $f(a)=0$. Hence, $f(0)=2$. Substitute $y=2$ into the original equation: $f(2x+f(x))=f(x+2)$. For any $x\neq0$, if $f(x+2)\neq0$, then $2x+f(x)=x+2$, which implies $f(x)=-x+2$. This also holds for $x=0$. Thus, $f(x)=-x+2$ holds for all $x$. In summary, either $f(x)\equiv0$, or $f(x)=x$, or $f(x)=-x+2$, are all valid solutions. They can be easily verified to meet the requirement.