How can I show that $uv = 5 + 2 i \sqrt{3}$ implies that either $u$ or $v$ is $1$ or $-1$?

Question

How can I show that $uv = 5 + 2 i \sqrt{3}$ implies that either $u$ or $v$ is $1$ or $-1$? Here $i = \sqrt{-1}$.

Answer 1

To rescue this problem, show
a = 5 + 2i$\sqrt 3$ is an irreducible Gausian integer.

If uv = a, then |u||v| = |uv| = |a| = 37.
As 37 is prime either |u| or |v| in {1,-1}.
Thus either u or v in {1,-1,i,-i}.

For example,a = ia(-i).

Answer 2

Assume to the contrary that both $u,\,v$ are some other nonzero complex numbers. In particular let $u=a+ib$ and $v=x+iy.$ Then we must have $$ax-by+i(ay+bx)=5+2i\sqrt 2.$$ Also, from the relation $|wz|=|w||z|,$ we have that $$(a^2+b^2)(x^2+y^2)=37=1+36=(ax-by)^2+(ay+bx)^2,$$ since the factors are positive and $37$ is prime. It follows that we must have that, for example, $ax-by=5$ and simultaneously, $ax-by=6,$ and now the proof is complete.