<s>[INST] 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.
 [/INST] 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 $f(x) = \sqrt{b}$.
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