Find $16$ complex numbers $x + yi, x,y\in \mathbb{Z}$ with norm $N(x + yi) =65$

Question

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.

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

Answer 1

$$65=5\times 13=|2+i|^2|3+2i|^2=|(2+i)(3+2i)|^2 =|4+7i|^2$$ $$65=5\times 13=|2+i|^2|3-2i|^2=|(2+i)(3-2i)|^2 =|8-i|^2$$ etc.

Answer 2

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$$\{ab\,|\,a\in A\wedge b\in B\}.$$

Answer 3

You have $65=8^2+1^2$ and also $65=7^2+4^2$. So you identify four numbers in the first quadrant as $1+8i, 4+7i, 7+4i, 8+i$ and proceed from there.