## A fact about irreducibles in ZZ[i]

Let $\pi = a+bi$ be an irreducible Gaussian integer with $a,b \neq 0$. Show that if $\pi$ is a factor of its conjugate $\overline{\pi} = a-bi$, then $\pi$ is an associate of $1+i$.

Let $\tau = c+di$ and suppose $\pi\tau = \overline{\pi}$. This equality yields the two equations $ac-bd=a$ and $ad+bc=-b$, which can be rearranged as $a(c-1) = bd$ and $ad = -b(c+1)$. Now $ad^2 = -bd(c+1) = -a(c+1)(c-1)$, so that \$latex $d^2 + c^2 = 1$. Hence $\tau \in \{ 1,-1,i,-i \}$.

If $\tau = 1$, then $a+bi = a-bi$, so that $b = 0$, a contradiction. Similarly, if $\tau = -1$ then $-a-bi = a-bi$ and we have $a = 0$. If $\tau = i$, then $-b+ai = a-bi$, so that $a = -b$. If $\tau = -i$, then $b-ai = a-bi$, and we have $a = b$. In either case, $|a| = |b|$. If $|a| > 1$, then $\pi = a+bi$ has a nontrivial factorization (namely $a(1+i)$) and so is not irreducible. Thus $|a| = 1$, and so $\pi \in \{1+i, 1-i, -1+i, -1-i\}$. These are precisely the associates of $1+i$.