Show that $7$ is irreducible in $\Bbb Z[i]$

Question

I have to show that $7$ is irreducible in $\Bbb Z[i]$.

To show irreducibility I have to show that it's not a unit. This is simple to just show exhaustively.

I'm having trouble with the second part which is to show that if it factors into $a.b$ that either $a$ or $b$ is a unit.

What I have so far is that

$7 = (a + bi)(c + di)$

$7 = (ac - bd) + (ad + cb)i$

Which gives two linear equations

$ad+cb=0$

$ac-bd=7$

How do I get from that to a complete proof? Or have I gone down the wrong path.

Answer 1

Hint: First, show that $a+bi$ is a unit if and only if $|a+bi|^2=1$ (that is, if and only if $a^2+b^2=1$). Next, show that $a+bi\mapsto|a+bi|^2$ is a multiplicative function, meaning that $|(a+bi)(c+di)|^2=|a+bi|^2|c+di|^2.$ When $7=(a+bi)(c+di),$ what can we conclude about the possible values of $|a+bi|^2$ and $|c+di|^2$? What does that tell us about $7$?