parmenides51 wrote: Find all functions $f : R \to R$ that satisfy the functional equation $$f(x + f(xy)) = f(x)(1 + y).$$ $\boxed{\text{S1 : }f(x)=0\quad\forall x}$ is a solution. So let us from now look only for non allzero solutions. Let $P(x,y)$ be the assertion $f(x+f(xy))=f(x)(1+y)$ Comparing $P(0,0)$ with $P(0,1)$, we get $f(0)=0$ Since non allzero, let $u\ne 0$ such that $f(u)\ne 0$ $P(u,\frac xu)$ $\implies$ $f(f(x)+u)=f(u)(1+\frac xu)$ and so $f(x)$ is bijective. $P(-x,-1)$ $\implies$ $f(f(x)-x)=0=f(0)$ and so, since injective, $\boxed{\text{S2 : }f(x)=x\quad\forall x}$, which indeed fits

The answers are $f(x)=x$ for all $x\in \mathbb{R}$ and $f(x)=0$ for all $x\in \mathbb{R}$. It’s easy to see that these functions satisfy the given equation. We now show these are the only solutions. Say there exists no $x\neq 0$ such that $f(x) \neq 0$. Then, $P(0,1)$ gives, \[f(f(0))=f(0) \implies f(r)=r\]which is clearly false if $r\neq 0$. Thus, the only solution in this case is $f\equiv 0$. Now, we assume that $f$ is not constant 0, then, there must exist some $x_0\neq 0$ such that $f(x_0) \neq 0$. Now, consider $\alpha,\beta \in \mathbb{R}$ such that $f(\alpha)=f(\beta)$. Then, $P\left(x_0,\frac{\alpha}{x_0}\right)$ and $P\left(x_0,\frac{\beta}{x_0}\right)$ yields, \[f(x_0)\left(\frac{\alpha}{x_0}+1\right) = f(x_0+f(\alpha))=f(x+f(\beta))=f(x_0)\left(\frac{\beta}{x_0}+1\right)\]which implies that $\alpha = \beta$ from which it is clear that $f$ is injective. Now, $P(x,0)$ gives, \[f(x+f(0))=f(x)\]which since $f$ is injective implies $f(0)=0$. Now, finally $P(-x,-1)$ gives, \[f(-x+f(x))=f(x)(0)=0=f(0)\]which since $f$ is injective implies that $f(x)=x$ for all $x\in \mathbb{R}$ which indeed shows that all solutions to the given functional equation are of the claimed forms.

For $P(0,0)$; $f(f(0))=f(0)$ $=====>$ $f(f(b))=f(b)$ Then for $P(0,1)$; $f(f(0))=f(0)*2=f(0)$ So, $f(0)=0$. $P(a; \frac{x}{a})$ =====> $f(a+f(x))=f(a)*(1+ \frac{x}{a})$ For $P(-x;-1)$; $f(f(x)-x)=f(-x)*0=0=f(0)$ $f(x)-x=0$ $f(x)=x$ Finally, we get two solutions; $f(x)=0$, $f(x)=x$

The only solutions are $f\equiv 0$ and $f(x) = x$, which work. Now we prove they are the only solutions. Let $P(x,y)$ be the given assertion. Clearly $0$ is the only constant solution, so assume $f$ nonconstant. Now choosing $x$ with $f(x) \ne 0$ and varying $y$ on the RHS gives $f$ is surjective since the RHS is always in the image of $f$. $P(1, y): f(1 + f(y)) = f(1) (1 + y)$. Claim: $f(1) \ne 0$. Proof: Suppose $f(1) = 0$ Then $f(f(y) + 1) = 0$ for all $y$. But then $f$ would be constant since it is surjective, absurd. Claim: $f$ is injective Proof: If $f(a) = f(b)$, then $P(1, a)$ compared with $P(1,b)$ gives that $f(1)(1 + a) = f(1)(1 + b)$, so $a = b$. $\square$ $P(x,0): f(x + f(0)) = f(x) \implies f(0) = 0$. $P(x, -1): f(x + f(-x)) = 0 = f(0)$, so $f(-x) = -x$, implying that $f$ is the identity.