If the norm of $\alpha \in \mathbb{Z}[i]$ is a square, show that $\alpha$ is a square in $\mathbb{Z}[i]$

Question

Suppose $\alpha \in \mathbb{Z}[i]$, $\alpha$ is not divisible by any integer, and $N(\alpha) = m^2, m \in \mathbb{Z}$.

I want to show that $\alpha$ is a square in $\mathbb{Z}[i]$.

I'm really lost and not sure how to proceed. I appreciate any hints.

Answer 1

Let $\alpha= a+ib$ with $a,b \in \mathbb Z$. We have $$(a+ib)(a-ib)=m^2$$ If $(a+ib)$ and $(a-ib)$ are coprime in $\mathbb Z$, they are both squares cause they build up a square : $m^2$. Suppose they are not coprime, there is some prime $\beta \in \mathbb Z[i]$ such that it divides both $(a+ib)$ and $(a-ib)$. $\overline \beta$ must be in the left hand side factorization cause the right hand side is an integer, but it can't be in $(a+ib)$ nor in $(a-ib)$, otherwise $\beta\overline \beta \in \mathbb Z$ divedes (a+ib) or $(a-ib)$.

Answer 2

Write $\alpha=a+bi$, then $N(\alpha)=a^2+b^2=m^2$, so $(a,b,m)$ is a Pythagoreen triple. This triple must be primitive, as by assumption $\gcd(a,b)=1$. Thus $a=x^2-y^2$, $b=2xy$, $m=x^2+y^2$, for some $x,y \in \Bbb Z$ (or the same equations with $a$ and $b$ switched.)

We get $a+bi=(x+yi)^2$

If $a=2xy$, and $b=x^2-y^2$, then $a+bi=-i(y+xi)^2$, this is at least a square up to a unit.