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

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.

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 $$a=2k+1\equiv -1\pmod{4}$$.