When the function equation $f(x)f(y)=axy+b$ is solvable 
Assume $a,b$ are constants. The question is whether there is a continuous function $f$ defined on $\mathbb R$ or $\mathbb C$ so that
$$
f(x)f(y)=axy+b
$$

Of course, such a function $f$ exists if $b=0$ by taking $$f(x)=\sqrt{a}x\,.$$  Likewise if $a=0$ then $f$ exists by taking $$f(x)=\sqrt{b}\,.$$ But I don't know whether the condition $a=0$ or $b=0$ is also necessary for the solvability of this function equation.
 A: Let's assume that $b \neq 0$, then clearly $f$ cannot be identically $0$. Thus, there exists $c$ such that $f(c)\neq 0$. Let $y=c$, then we get $f(x)f(c)=acx+b$, or after division $f(x)=\frac{ac}{f(c)}x+\frac{b}{f(c)}$, and so $f(x)$ is linear function. So, let $f(x)=px+q$ and plug it back to the original equation. We get
$$(px+q)(py+q)=axy+b$$
for all $x,y$, so after expanding and comparing the coefficients, we get $p^2=a,pq=0,q^2=b$. By original assumption, $q \neq 0$, and so $p=0$, $a=0$, and $q=\sqrt{b}$. So the only solution then is $f(x)=\sqrt{b}$, which works only when $a=0$.
A: $f(0)f(y) = b$ suggests $f(z) = \sqrt{b}$ is another solution without the requirement $b = 0$.  (First, it suggests $f(y)$ is independent of $y$, so constant.  Then we have the product of two copies of this constant is $b$.)
A: If $x$ and $y$ are independent variables, you only have to set $y = x$, to get $f(x)^2 = ax^2 + b$ and therefore $f(x) = \sqrt{ax^2 + b}$. Now,
$$f(x)f(y) = \sqrt{(ax^2 + b)(ay^2 + b)} = \sqrt{a^2x^2y^2 + abx^2 + aby^2 + b^2}$$
Needs to be equal to $axy + b$, i.e. taking the squares of both expressions
$$a^2x^2y^2 + abx^2 + aby^2 + b^2 = (axy + b)^2 = a^2x^2y^2 + 2abxy + b^2$$
which entails $a = 0$ or $b = 0$. Thus it is either $f(x) = x\sqrt{a}$ or $f(x) = \sqrt{b}$.
