Find all function $f:\ \Bbb{R}\ \longrightarrow\ \Bbb{R}$ such that : $f(ax)f(by)=f(ax+by)+cxy$ where $a,b,c>0$

Question

If $f:\ \Bbb{R}\ \longrightarrow\ \Bbb{R}$ and $a,b,c>0$, then find all function such that : $$f(ax)f(by)=f(ax+by)+cxy,\quad \text{where } a,b,c>0 \text{ for all } x,y\in \Bbb{R}.$$ My attempt

• When $x=0$ and $y=0$, we find $f(0)=1$ or $0$
• If $f(0)=1$ then take $x=0$, we find $f(by)=f(by)$

I don't know how I complete and get answer!!

Help me or hint me please. Thanks!

Answer 1

The functional equation looks a bit simpler when you substitute $u:=ax$, $v:=by$ and $d:=\tfrac{c}{ab}>0$; the functional equation then becomes $$f(u)f(v)=f(u+v)+duv.$$

As you already note, plugging in $u=v=0$ shows that $$f(0)f(0)=f(0)+d\cdot0,$$ and so $f(0)\in\{0,1\}$. If $f(0)=0$ then plugging in $u=0$ shows that $$f(0)f(v)=f(v),$$ for all $v$, and hence that $f=0$. Otherwise $f(0)=1$ and then plugging in $v=-u$ yields $$f(u)f(-u)=f(0)-du^2=1-du^2.$$ Can you continue from here?