Find $16$ complex numbers $x + yi, x,y\in \mathbb{Z}$ with norm $N(x + yi) =65$ I took some integer value for $x,y$, as $x=3, y=2$ with norm  $N = \sqrt{13}$. But, no it is not. I  hoped that by finding a relation between current value of $x,y, N$ to the desired one; would be able to get the desired $x',y'$. But could not find any such relation, as :
So, tried to square the values of $x,y$ to get $x'=x^2=9, y'=y^2=4$; as hoped that it will have norm $13$. 
Definitely, the reason lies in adding $2x'\cdot y'$ also, whose square root need be split in $x',y'$.
To check this idea, $x'^2= x^4 = 81; y'^2 = y^4 = 16; 2x'y'= 72$ and $x'^2 + y'^2 + 2x'\cdot y' = 81+16+72 =  169$.
The square root of $72$ is not an integer, so cannot go further.
So, for getting a norm of $65$ cannot proceed from norm of $\sqrt{13}$.
Could not get any other way out. Please tell a logical approach to generate all such pairs.

Addendum Sorry, for flawed defn. of norm. Total question is created out of that wrong definition.

Let$$A=\{\pm(2+3i),\pm(2-3i),\pm(3+2i),\pm(3-2i)\}$$and$$B=\{\pm(1+2i),\pm(1-2i),\pm(2+i),\pm(2-i)\}.$$Now, consider the set$$\{ab\,|\,a\in A\wedge b\in B\}.$$
