Simple proof involving functional equations

Question

I'm struggling with a proof involving a simple functional equation. The problem asks to show that if $f$ satisfies ( for all $x,y$ real): $$f(xy+x+y) = f(xy)+f(x)+f(y),$$ then $f$ must also satisfy $$f(x+y) = f(x)+f(y).$$

I tried contrapositive. Suppose there is one pair $(x,y)$ that satisfies the functional equation, but for which the second equation is not satisfied. Then, it follows that $$f(xy+x+y) \ne f(xy)+f(x+y).$$ That is, if $(x,y)$ does not hold for the second equation, then neither does $(xy,x+y)$. This gives us a collection of points that, if the functional equation is true, also do not hold for second equation. My goal was to show that the second equation would fail at all points and thus give a contradiction (because both equations are clearly satisfied if $x = 0$, $y = 0$, for example), but not sure if this is the right approach.

I'm new to study of functional equations, so bear with me.

Answer 1

Here is my proof (watch out for spoilers: it's a complete proof, not just a hint!):

Let's assume that $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfies $$f(xy+x+y) = f(xy)+f(x)+f(y)$$ for every pair of real numbers $x,y$.

1. $f(x) = -f(-x)$, in particular $f(0) = 0$
   Trivial, just plug $y=-x$ into the original functional equation.
2. $f(x+xy+y) + f(x - xy -y) = 2f(x)$
   By plugging $-y$ instead of $y$ into the original functional equation we get:$$f(x - xy -y) = f(-xy) + f(x) + f(-y) = -f(xy) + f(x) - f(y) $$ Now add this equation to the original functional equation.
3. $f(x+y) + f(x-y) = 2f(x)$ for $x \neq -1$
   Just plug $y/(x+1)$ instead of $y$ into the functional equation from point 2.
4. $f(x+y) + f(x-y) = 2f(x)$ for all $x$
   From 3. (plugging $x = 1$) we have $f(1+y) + f(1-y) = 2f(1)$. Thus also $f(-1-y) + f(-1+y) = 2f(-1)$.
5. $f(2x) = 2f(x)$
   We get this from the functional equation in 4. by plugging $y=x$.
6. $f(x) + f(y) = 2f(\frac{x+y}{2})$
   We get this from the functional equation in 4. by plugging $\frac{x+y}{2}$ instead of $x$ and $\frac{x-y}{2}$ instead of $y$.

Now combine 6. and 5. to get the thesis.