Answer
Yes, if the use of imaginary number $i$ is allowed..
An example is $4x^2+9y^2=(2x-3yi)(2x+3yi)$.
Work Step by Step
We can factor the sum of two squares. However, as we will see, there will be an imaginary number $i$ in the solution. ($i = \sqrt {-1}$)
Also, we can only factor the sum of two squares when both terms are squares.
Example:
$$4x^2+9y^2=(2x)^2+(3y)^2(1)$$
Since $1 = -1\cdot -1$, we can rewrite the previous line as
$$\begin{align*}
4x^2+9y^2&=(2x)^2+(3y)^2(-1)(-1)\\
&=(2x)^2-(3y)^2(-1)
\end{align*}$$
Since $\sqrt{-1}\cdot \sqrt{-1} = -1$ and $\sqrt{-1}=i$, we can rewrite the previous line as:
$$\begin{align*}
4x^2+9y^2&=(2x)^2-(3y)^2(\sqrt{-1})(\sqrt{-1})\\
&=(2x)^2-(3y)^2(i)(i)\\
&=(2x)^2-(3y)^2(i^2)\\
&=(2x)^2-(3yi)^2
\end{align*}$$
Using the formula $a^2-b^2=(a-b)(a+b)$, the expression above can be factored as:
$$4x^2+9y^2=(2x-3yi)(2x+3yi)$$
Thus, $4x^2+9y^2=(2x-3yi)\cdot (2x+3yi)$.
