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

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!

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?
• Try finding some more identities for $f$ by plugging in more values for $u$ and $v$. – Servaes Apr 25 at 8:39