4. Find two numbers, such that the product of their squares is 5500, and the product of their difference and the square of the latter is 352.
Let $x, y$ represent the two numbers we are looking for, then
$$
\begin{array}{l}
(x+y)\left(x^{2}+y^{2}\right)=5500, \\
(x-y)\left(x^{2}-y^{2}\right)=352 .
\end{array}
$$

Therefore, $\frac{(x+y)\left(x^{2}+y^{2}\right)}{(x-y)\left(x^{2}-y^{2}\right)}=\frac{5500}{352}$,
which simplifies to $\frac{x^{2}+y^{2}}{(x-y)^{2}}=\frac{125}{8}$.
Thus, $117 x^{2}-250 x y+117 y^{2}=0$,
or $(13 x-9 y)(9 x-13 y)=0$.
Hence, $\frac{x}{13}=\frac{y}{9}=k$.
Substituting into the second equation yields
$$
352=4 k \times 88 k^{2} .
$$

So, $k=1$.
Therefore, the two numbers are $x=13, y=9$. 

(Note: The original solution had a typo stating $y=0$, which should be $y=9$ based on the steps provided.)