I want to show that $(a + bi)|(c + di)$ is equivalent to the statement that the ordinary integers $(a^2 + b^2)|(ac + bd)$ and $(a^2 + b^2)|(-ad + bc)$.
I also want to show that $(a + bi)|(c + di) \implies (a^2 + b^2) | (c^2 + d^2)$
So for the first part do I just multiply two complex numbers together then square it? Can this idea be used for the other parts of this problem as well? The help would be appreciated!
On
Hint $\ \ \overbrace{\bar \alpha\mid \bar\beta \iff \alpha\mid \beta}^{\rm multiply\ for\ part\ 2} \iff \overbrace{\alpha\bar\alpha\mid \beta\bar\alpha}^{\rm part\ 1}\ $ if $\ \alpha \neq 0$
Remark $\ $ Part $1$ is a generalization of *rationalizing the denominator, see here for more.
No: what it wants you to do (presumably) for the first one is to multiply both sides of the division by $(a-bi)$. (This is fine provided $a-bi \neq 0$: if $p \neq 0$ then $m\mid n \iff pm \mid pn $.) For the second one, you can work out from the final form a similar expression to multiply by.
For the third one, use $m \mid n $ and $p \mid q \implies mp \mid nq $.