Help working with the functional equation $f(a) f(b) = f(a+b) + ab$

Question

I have looked at some of the other problems on here, but am stuck. Here is the problem: Consider $f: \mathbb{R} \to \mathbb{R} $ such that $f(a) f(b) = f(a+b) + ab$ for all real $a$ and $b$: find all possible $f$.

I tried:

• Let $a=b=0$ to get $f(0)f(0) = f(0) + 0$. Then I factored, so $f(0)=0 $ or $f(0) = 1$
• Then, I let $a = 0$, ie $f(0)f(b) = f(b) + 0$. I think this implies that $f(0) = 1$.

I am not sure how to proceed. More substitutions until I determine a pattern? I know I need to talk about something in terms of a constant eventually, but cannot find a way to get to that point.

Answer 1

From $f(0)=1$, we get that $f(1)f(-1)=0$:

• $f(1)=0$ implies $f(b+1)+b=0$ that is $f(x)=1-x$. Let's check: $(1-a)(1-b)=1-(a+b)+ab$: True!
• $f(-1)=0$ implies $f(b-1)-b=0$ that is $f(x)=x+1$. Let's check: $(a+1)(b+1)=(a+b)+1+ab$: True!