Functions satisfying $f(x+y)+x= f(x)f(y)$

Question

Let $f$ be a real valued function such that $f(x+y)+x=f(x)f(y)$, for every $x$ and $y$.

I have tried $x=0$, it seems the exponential function works, but for $x\neq 0$ and $y\neq 0$ I can't succed to get functions satisfying $f(x+y)+x=f(x)f(y)$.

Answer 1

There are no such functions. To see that, suppose we had such a function. We will derive a contradiction.

First note that the constant function $0$ does not work, hence there must be some value $y_0$ for which $f(y_0)\neq 0$.

Taking $x=0, y=y_0$ then yields $$f(y_0)=f(0)\,f(y_0)\implies f(0)=1$$

Now, taking $y=0$, with no restriction on $x$, we get $$f(x)+x=f(x)$$ which is absurd.

Answer 2

If $x=y =0$, it gives

$$f (0)=f (0)^2$$ thus $f (0)=0$ or $f (0)=1$.

Assume $f (0)=0$.

with $x=-y $, we get $$f (0)+x=f (x)f (-x)=-x $$ which is not possible.

so, $f (0)=1$.

with $x=-y $, $$1+x=f (x)f (-x)=1-x $$ There is no function satsfying that condition.

Answer 3

$x = 0, y = 1$, $$f(1) + 0 = f(1)f(0)$$

$x = 1, y = 0$, $$f(1) + 1 = f(1)f(0)$$

This implies $0 = 1$ so no such function exits.