# Chinese remainder theorem (Gaussian ints)

1. Nov 23, 2009

### math_grl

Figured it would be a quick review of ideas that turned into a nightmare...

$$x \equiv i ($$mod $$1 + i)$$
$$x \equiv 1 ($$mod $$2-i)$$
$$x \equiv 1+i ($$mod $$3+4i)$$

we find $$s \equiv 0$$(mod $$1+i)$$ and $$s \equiv 1$$(mod $$5+10i)$$ but this is easy.

So now find $$s' \equiv 0$$(mod $$2-i)$$ and $$s' \equiv 1$$(mod $$-1+7i)$$ is complicated...Using the euclidean algorithm I got that there is a gcd of $$\pm i$$ for $$2-i$$ and $$-1+7i$$

I think i'm kind of stuck, I also can't seem to find
$$s'' \equiv 0$$(mod $$3+4i)$$ and $$s'' \equiv 1$$(mod $$3+i)$$.