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We have $a, b \in \mathbb{Z}[i]$

$a=4$

$b=3+3i$

Then, are there any $u, v \in \mathbb{Z}[i]$ such that $ua+vb=−1+5i$?

Fernando
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  • What have you tried? – Fimpellizzeri Jan 10 '20 at 16:51
  • In general, you can use Euclid's algorithm. If there is a solution, then $\frac{(-1+5i)-vb}{a}\in\mathbb{Z}[i]$. Now, compute the remainders $r_1,r_2$ of the division of $-1+5i$ and $b$ by $a$. Then there is $u_1\in\mathbb{Z}[i]$ such that $au_1=r_1-vr_2$. Now, repeat but this time with $r_2$: We must have $\frac{r_1-au_1}{r_2}\in\mathbb{Z}[i]$. Compute the remainders $r_3,r_4$ of $r_1$ and $a$ divided by $r_2$. Then there is $v_1\in\mathbb{Z}[i]$ such that $r_2v_1=r_3-r_4u_1$. Eventually you get to either a solution or a constant quotient that is not in $\mathbb{Z}[i]$. – MoonLightSyzygy Jan 10 '20 at 16:56
  • @DonThousand: OP asked for $u,v\in\mathbb{ \color{red}Z[i]}$ – J. W. Tanner Jan 10 '20 at 16:56
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    @J.W.Tanner I clearly don't know how to read, lmao. – Rushabh Mehta Jan 10 '20 at 16:57
  • Split into real part and imaginary part, solve simultaneous equations. – Sam Jan 10 '20 at 17:10

2 Answers2

1

Hint:

$a=(1+i)(2-2i)$

$b=(1+i)3$

J. W. Tanner
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0

Hint: with the same proof as in $\Bbb Z$ (or any PID), $\,\exists\, x,y\!:\ ax+by = c\iff \gcd(a,b)\mid c.\ $ Recall that $\,\Bbb Z[i]\,$ has a constructive (norm) Euclidean algorithm so we can always quickly compute this gcd, which determines solvabilty (and also the general solution - by scaling - follow the link).

Or: $\,(4,3(1\!+\!i)) = (4,i\!+\!1) = 1\!+\!i,\ $ by $\ 1\!+\!i\mid j\!+\!ki\iff 2\mid j\!-\!k\ $ (so $\ 1\!+\!i\mid -1\!+\!5i),\,$ thus

$$\begin{align} &4x + (3\!+\!3i) y \,=\, j\! +\! ki\ \ {\rm is\ solvable}\\[.2em] \iff\ &1\!+\!i\mid j\!+\!ki\ \ \ {\rm by}\ \ \ 1\!+\!i = \gcd(4,\,3\!+\!3i)\\[.2em] \iff\ &\ \ \ \ \ 2\mid j\!-\!k\\[.2em] \iff\ &\ \ \ \ \ \ j,\,k\ \,\text{have equal parity}\end{align}\qquad$$

This is equivalent to saying that $\,j\!+\!ki\,$ is "even" in $\,\Bbb Z[i],\,$ as is explained in this post on notions of parity for algebraic numbers.

Bill Dubuque
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