Congruence proof. Show $a$ is odd, $b$ is even, and $ a \equiv 1$ mod $4 $ under certain conditions.

Question

If $ p \equiv$ $1$ mod $4$, $ p = a^2 + b^2$, and $a + bi \equiv 1$ mod $2+2i$, then $a$ is odd and $b$ is even. Moreover, if $4|b$, then $ a \equiv 1$ mod $4$, and if $ 4 \nmid b$, then $ a \equiv -1$ mod $4$.

Proof: $ a + bi \equiv 1(2+2i) $ implies that $ a + bi \equiv 1$ mod $2 $. and so $a$ is odd and $b$ is even. Since $ 4 = -2(i-1)(i+1) $ it follows that if $4 \mid b$ then $a+bi \equiv a \equiv 1$ mod $2+2i$. Taking conjugates $ a \equiv 1$ mod $2-2i $. Thus $(2+2i)(2-2i) = 8 \mid (a+1)^2 $ and $ a \equiv 1$ mod $4$. If $ 4 \nmid b $ then $ b = 4k + 2 $ for some $ k $. Thus $ a + bi \equiv a+2i \equiv 1$ mod $2+2i $. Since $ 2i \equiv -2$ mod $2+2i $ we have $ a \equiv 3 \equiv -1$ mod $2+2i $. As before $ 8 \mid (a+1)^2 $ and so $ a \equiv -1$ mod $4 $.

So I'm trying to understand this proof, but unfortunately I'm a beginner.So I don't understand this proof. Here are my questions:

1) Why implies $ a + bi \equiv 1(2+2i) $ that $ a + bi \equiv 1$ mod $2 $? And why is this enough to say that $a$ is odd and $b$ is even?

2) Why do we have $a+bi \equiv a \equiv 1$ mod $2+2i$, if $4|b$? ( second line of the proof )

3) Why do we have $(2+2i)(2-2i) = 8 \mid (a+1)^2 $ and $ a \equiv 1$ mod $4$ ?

I can imagine that these questions are very simple for you, but I really want to understand this proof. So please try to be accurate. Thank you.

Answer 1

First, $a+bi \equiv 1\pmod{2+2i}$ means that $a+bi = 1 + k(2+2i) = 1+2k(1+i)$ for some integer $k$, so that $a+bi \equiv 1\pmod{2}$. As for your second question, $4 = -2(i-1)(i+1) = (1-i)(2+2i)$, so $4\mid b$ implies that $2+2i\mid b$ and thus $bi\equiv 0\pmod{2+2i}$. For your final question, I think the correct statement is $8\mid (a-1)^2$, not $(a+1)^2$: since $2+2i\mid a-1$ and $2-2i\mid a-1$, this is clear.

Here's another proof that you may find easier to understand: $a+bi\equiv 1\pmod{2+2i}$ means that $a+bi = 1+k(2+2i) = 1+2k(1+i)$ for some integer $k$. Rewriting as $a+bi = (2k+1)+2ki$ shows that $a$ is odd and $b$ is even. Now, if $4\mid b$, then $k$ must be even, so that $a=2k+1\equiv 1\pmod{4}$, while if $4\nmid b$, then $k$ must be odd, so that $a=2k+1\equiv -1\pmod{4}$.