Conditions on the functions $f,g,h,k$ if $f(x)g(y)=h(x)k(y)$ I was working on this problem, and I thought I'd post my answer so people could see if they have a better one:
Spivak Calculus, 4th ed., problem 3-18:
Suppose $f,\,g,\,h,\,k$ are functions from $\mathbb{R}$ to $\mathbb{R}$. Precisely what conditions must $f,\,g,\,h,\,$ and $k$ satisfy in order that 
$$
f(x)g(y)=h(x)k(y) \tag{1}
$$ 
for all $x,\,y\,\in \mathbb{R}$?
 A: Condition (1) is equivalent to saying that the matrix 
$$A(x,y)=\begin{pmatrix} f(x) & h(x) \\ k(y) & g(y) \end{pmatrix}$$
is singular for all $x,y$. This is equivalent to its columns being linearly dependent: i.e., for all $x,y$ there exist $a,b$ (not both zero) such that 
$$af(x)+bh(x) =0 = ak(y)+bg(y)$$
In principle, $a $ and $b$ could depend on $x$ and $y$.  To clarify the matter, consider two cases.


*

*One of the rows of $A$ is identically zero. Then there are no   conditions on the other row; property holds.

*Both rows are  sometimes nonzero. Then we can choose $a,b$ above independent of $x,y$. Indeed, fix $x,y$ such that $A(x,y)$ has nonzero rows, and find $a,b$ for this matrix. The same $a,b$ must work for every other $A(x',y' )$, as we can see by considering $A(x',y)$ and $A(x,y')$ first.


Summary: (1) holds if and only if one of the following is true:


*

*$f\equiv 0 \equiv h$; 

*$g\equiv 0 \equiv k$; 

*There exist constants $a,b$, not both zero, such that $af+bh\equiv 0\equiv ak+bg$.

A: First of all there is the possibility that both sides of (1) are identically zero. For this it is necessary that 
$$
\text{One of $f,\,g$ and one of $h,\, k$ are identically zero}
$$
since if not, we can find $x,y$ such that one side is nonzero.
If neither side is identically zero, then let $Z_f=\{x \in \mathbb{R}\mid f(x)=0\}$, and similarly $Z_g,\,Z_h,\,Z_k$.
Observe: If neither side of (1) is identically zero, we have 
$$
Z_f = Z_h \quad \text{ and } \quad Z_g=Z_k,
$$
and 
$$
\frac{f}{h} \text{ is constant on $(Z_f)^c$ and } \frac{k}{g} \text{ is constant on $(Z_k)^c$, and they are the same constant.}
$$
This seems to be the most we can say. Anyone see anything else?
