Finding the Sum of the square of two positive integers. 
Write the following equation as the sum of the square of two integers, $a^2 +b^2$.
  $$(8^2+5^2)(13^2+7^2)$$

I remember that you are supposed to do something with complex numbers or at least that is what my teacher did in class.
 A: If you really want the complex number method for tackling this, you can use the factorisation $a^2+b^2 = (a+ib)(a-ib)$ and then do this:
$$(8^2+5^2)(13^2+7^2) = (8+5i)(8-5i)(13+7i)(13-7i)$$
and then group the brackets differently:
$$
\begin{align}
& = [(8+5i)(13+7i)] . [(8-5i)(13-7i)]\\
& = (69+121i)(69-121i)\\
& = 69^2+121^2
\end{align}
$$
There is another possible answer you can get by grouping the brackets differently, I will leave you to find it.
A: From identity $(a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-bc)^2$, that is easy to check, we get:
$$(8^2+5^2)(13^2+7^2)=(8\cdot13+5\cdot7)^2+(8\cdot7-5\cdot13)^2=139^2+9^2.$$
A: Hint $\quad\ \begin{eqnarray}\alpha = a+bi\\ \\ \beta = c+di\end{eqnarray} \quad \Rightarrow\quad  $ $\begin{eqnarray} \color{#c00} {(\alpha \bar\alpha)}&&\ \ \ \, \color{#c00}{(\beta\bar\beta)} &\,=\,& \quad\ \color{#c00}{\alpha\beta\  \overline{(\alpha\beta)}}\\
\color{#c00}{N(\alpha)}&&\color{#c00}{N(\beta)} &=& \color{#c00}{N(\alpha\beta)},\quad\ N := \rm norm\\
N(a\,+\,b\,i)&&\!N(c\,+\,d\,i) &\,=\,& N(ac\!-\!bd\ +\ (ad\!+\!bc)\,i) \\
(a^2+b^2)\ &&\ \ \, (c^2\!+d^2) &\,=\,&\ \ \ (ac\!-\!bd)^2\! + (ad\!+\!bc)^2 \end{eqnarray}$
