I don't know how to use latex very well

mijail wrote:

I don't know how to use latex very well Without proving surjectivity how can you told that $f(a)=0$ P.S. If you find $f(some value)=0$ than $f(c)=0$ is well. Some value=$f(1)$,$f(0)$ etc. Please elaborate your idea.

MiltonMath12 already gave a full solution.

We claim that the only solutions that work are $f(x)=x+1$ for all $x$ and the constant solution $f(x)=0$ for all $x$. Denote the assertion by $P(x,y)$. $$P(0,f(0)) \implies f(0)f(-f(0))=0 \implies f(0)=0 \quad \text {or} \quad f(-f(0))=0$$ Assume $f(0)=0$. Then we have : $P(0,-y)\implies f(y)=0.$ So this gives us $f(x)=0$ for all $x$ which works. Henceforth assume $f(-f(0))=0$. We will assume that $f$ is non-constant because the only constant solution is $f\equiv 0$ which has been dealt with. Next let $a=-f(0)$. We have : $$P(a,1)\implies f(-1)=0$$$$P(x,1) \implies f(f(x)-1)= f(x)$$ Now we will prove that $f$ is injective. Assume that there exists $a\neq b$ with $f(a)=f(b)$ with $ab\neq 0$. We have by $P(a,y)$ and $P(b,y)$ that $f(ay)=f(by)$ for all real $y$. We now define the set $S$ as follows : $$ S := \{s\in S : f(sx)=f(x) \, \forall x\in \mathbb R\}$$ Note that $s\in S \iff \tfrac 1s\in S$. We prove that if $f$ is not injective then $S \equiv \mathbb R$, which proves $f$ is constant (contradiction). $P(x,y)$ and $P(x,sy)$ with $s<1$ and $s\in S$, gives: $$f(f(x)-y))=f(f(x)-sy) \implies \frac {f(x)-sy}{f(x)-y} \in S$$ Note that as we vary $y$, $\tfrac {f(x)-sy}{f(x)-y}=1+ \tfrac {y(1-s)}{f(x)-y} $ sweeps all reals, so the claim holds. Hence we conclude $f$ is injective. So we have : $$f(f(x)-1)=f(x)\implies \boxed {f(x)=x+1}$$ As desired. $\blacksquare$

Let $P(x,y)$ be the assertion $f(f(x) - y) = f(xy) + f(x)f(-y)$ Obviously $f\equiv 0$ is the only constant solution. $P(0,f(0))$ and $P(-f(0),1)$ give us $f(-1)=0$. $P\left(x,\frac{f(x)}{x+1}\right)\implies f(x)f\left(-\frac{f(x)}{x+1}\right)=0$ $(\heartsuit)$ Suppose there exists $u\neq -1$ such that $f(u)=0$. $P(u,y)\implies f(-y)=f(uy)$ or $f(x)=f(-ux)\forall x\in\mathbb R$. $P(x,uy)$ with $P(x,-y)$ implies $f(f(x)+y)=f(f(x)-uy)$ so if $u\neq 1$ we have that $f$ is constant, contradiction. So $f(u)=0\iff u=-1$. So using $(\heartsuit)$ this means $f\left(-\frac{f(x)}{x+1}\right)=0$ thus $-\left(\frac{f(x)}{x+1}\right)=-1$ which means $f(x)=x+1$.

Nice wrapped FE! Let $P(x,y)$ be the assertion $f(f(x) - y) = f(xy) + f(x)f(-y)$. Observe that $f \equiv 0$ is the only constant solution. Now, suppose that $f$ is non constant. We claim that $f \equiv x+1$ is the only solution. Firstly, observe that $P(0,f(0))$ gives us that $f(0)f(-f(0))=0$. If $f(0)=0$, from $P(0,y)$ we have that $f(-y)=0$ for all $y \in \mathbb{R}$, which implies that $f$ is constant, a contradiction. Hence, $f(-f(0))=0$. $(\spadesuit)$ We prove that $f$ is injective. Observe that if $f(a)=f(b)$ for some $a,b \in \mathbb{R}$ such that $a \neq b$, then $P(a,y)-P(b,y)$ tells us that $f(ay)=f(by)$ for all $y$. $(\star)$ As Aryan-23, we define the set $$ S := \{s\in S : f(sx)=f(x) \, \forall x\in \mathbb R\}$$ Assume that $f$ is not injective, so $S$ contains an element $s$ distinct from $1$. Then, $P(x,ys)-P(x,y)$ gives that $f(f(x)-y)=f(f(x)-ys)$ for all $x,y$. On the other hand, we already know that $f(f(x)-y)=f(sf(x)-ys)$, so $f(sf(x)-ys)=f(f(x)-ys)=f(f(x)-ys+(s-1)f(x)$ for all $x,y$. Picking $x=0$, we have that since $s \neq 1, f(0) \neq 0$, $f(0)-ys$ sweeps all reals when we vary $y$ and $f(f(0)-ys)=f((s-1)f(0)+(f(0)-ys))$, we have that $f$ is periodic, i.e., there is a non-zero $c \in \mathbb{R}$ such that $f(x)=f(x+c)$ for all $x \in \mathbb{R}$. From $(\star)$, it is clear that if $f(a)=f(b)$ then $\frac{a}{b}, \frac{b}{a} \in S$. Thus, we have that $\frac{x+c}{x}=1+\frac{c}{x} \in S$ for all non-zero $x \in \mathbb{R}$. Since $1+\frac{c}{x}$ sweeps all reals, we have that $S \equiv \mathbb{R}$. This proves that $f$ is constant, a contradiction. Therefore, $f$ is injective. Now, from $(\spadesuit)$, $P(-f(0),y)$ gives $f(-y)=f(-f(0)y)$, so $y=yf(0)$ for all $y$, so $f(0)=1$. Moreover, $P(x,f(0))$ tells us that $f(f(x)-1)=f(x)$, so $f(x)=x+1$ for all $x \in \mathbb{R}$, which clearly works. $\blacksquare$

plagueis wrote: Find all functions $f \colon \mathbb{R} \to \mathbb{R}$ such that \[f(f(x) - y) = f(xy) + f(x)f(-y)\]for any two real numbers $x, y$. Proposed by Pablo Valeriano $P(f(a)-b,y)$ gives:$f(f(f(a)-b)-y)=f(y(f(a)-b))+f(f(a)-b)f(-y)\Rightarrow $ $f(f(ab)+f(a)f(-b)-y)=f(y(f(a)-b))+f(ab)f(-y)+f(a)f(-b)f(-y)$ In this for $b\rightarrow -b$ we get: $ f(f(-ab)+f(a)f(b)-y)=f(y(f(a)+b))+f(-ab)f(-y)+f(a)f(b)f(-y)$ Now by symmetry of $a,b$ we have that:$ f(y(f(b)+a))=f(y(f(a)+b))$ Taking to this $b=-f(a)$ we have that $f(y[f(-f(a))-a])=f(0)$ So if there is a $a$ such that $f(-f(a))\neq a$ we get that $f$ is constan and so $f(x)=0$. If $f(-f(a))=-a$ for every $a$ then we have that $f$ is epi and $1-1$. $P(x,0)$ gives: $f(f(x))=f(0)+f(x)f(0)\Rightarrow f(x)=cx+c$ And esily we have $f(x)=x+1$