Solve Equations in the ring of Gaussian Integers

Question

How to solve equations in Gaussian integers? For example: $$(7-i)x+(12-i)y = 2+3i.$$ Why can't I just rewrite it as $7x+12y = 2; -x-y = 3$? However, in that case the solution isn't integer.

Answer 1

If $x,y\in\mathbb{Z}$ then you can write what you have written, but remember that $x,y\in\mathbb{Z}[i]$ here. You can however let $x=u+vi, y=z+wi$ where $u,v,z,w\in\mathbb{Z}$ but this does not make it very much easier.

---

As have already become clear, the equation $$(7-i)x+(12-i)y=2+3i\tag{1}$$ has no solution for $x,y\in\mathbb{Z}[i]$. One way to see this is to calculate the norm of the coefficients. $$N(7-i)=7^2+(-1)^2=50\tag{2}$$ $$N(12-i)=12^2+(-1)^2=145\tag{3}$$ Notice that $\gcd(145,50)=5$, this means that $5=(2-i)(2+i)$ shares a factor with both $7-i,12-i$. From a little algebra we find that $$7-i=(3+i)(2-i)\tag{4}$$ $$12-i=(5+2i)(2-i)\tag{5}$$ Thus we may write eq. $(1)$ as $$(2-i)\left ((3+i)x+(5+2i)y\right )=2+3i\tag{6}$$ Eq. $(6)$ tells us that $2+3i$ is composite, this is however impossible since $N(2+3i)=2^2+3^2=13$ is a prime number, which makes $2+3i$ an irreducible.

Answer 2

Let $x=a+bi$ and $y=c+di$. Then the equation is equivalent to two equations over the integers, namely, $$ - a + 7b - c + 12d - 3=0,\; 7a + b + 12c + d - 2=0. $$ Substituting $c=7b-a+12d-3$ into the second equation, we obtain $$ - 5a + 85b + 145d - 38=0. $$ This is a contradiction modulo $5$. So there are no solutions.

Answer 3

$\mathbb{Z}[i]$ is a Euclidean domain, so you can solve problems like this in exactly the same way you would solve problems such as

Find all integers $x,y$ such that $63 x + 15 y = 2$

The problem you're given actually has a simple solution method the same as this example problem.

In this example, $\gcd(63, 15) = 3$, so we can probably either simplify the problem or learn something by considering division by 3 or equivalence modulo 3.

In this case, equivalence modulo 3 is informative. The equation $$ 63 x + 15 y \equiv 2 \pmod 3 $$ simplifes to $$ 0 x + 0y \equiv 2 \pmod 3 $$ which has no solutions. Thus the original equation $63 x + 15 y = 2$ can't have any integer solutions either.

An equivalent argument in this example is that $3$ divides the left hand side but not the right hand side, so the equation is a contradiction: it has no solutions.