Prime factorization of Gaussian integers

Question

I want to find $a, b\in\mathbb{Z}[i]$ such that $a(2+3i)+b(5+5i)=1$.

I don't know how to do this, but my first thought was to do something with the norm or otherwise factoring $5+5i=(2+i)(2-i)(1+i)$, but I don't see how to use this further.

The second question is for which $x\in\mathbb{Z}[i]$ we have $x=1\bmod (2-+3i)$ and $x=-1\bmod(5+5i)$? But I would like to try and solve this myself as soon as I have the first question.

Edit: I will attempt to solve this using the Euclidean algorithm as stated in the comments. I will be back later with more! (Although I would not mind if someone beat me to it) ;)

Answer 1

Using the Extended Euclidean Algorithm as implemented in this answer, modified for Gaussian integers, we get $$ \begin{array}{r} &&2&-1+2i&1-i\\\hline 1&0&1&1-2i&2+3i\\ 0&1&-2&-1+4i&-5-5i\\ 5+5i&2+3i&1-i&1&0 \end{array}\tag{1} $$ This says that $$ (1-2i+(2+3i)k)\color{#C00000}{(5+5i)}+(-1+4i-(5+5i)k)\color{#C00000}{(2+3i)}=1\tag{2} $$ where $k$ is a Gaussian integer. For example, $k=i$ gives the solution $$ \underbrace{(-2)\color{#C00000}{(5+5i)}}_{-10-10i}+\underbrace{(4-i)\color{#C00000}{(2+3i)}}_{11+10i}=1\tag{3} $$

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Since $-10-10i=(-2)(5+5i)$, we have $$ \begin{align} -10-10i&\equiv1\pmod{2+3i}\\ -10-10i&\equiv0\pmod{5+5i} \end{align}\tag{4} $$ and since $11+10i=(4-i)(2+3i)$ $$ \begin{align} 11+10i&\equiv0\pmod{2+3i}\\ 11+10i&\equiv1\pmod{5+5i} \end{align}\tag{5} $$ Adding $a$ times $(4)$ and $b$ times $(5)$ gives a solution to $$ \begin{align} x&\equiv a\pmod{2+3i}\\ x&\equiv b\pmod{5+5i} \end{align}\tag{6} $$

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Applying $(6)$ to the particular question above $$ \begin{align} -21-20i&\equiv \phantom{-}1\pmod{2+3i}\\ -21-20i&\equiv -1\pmod{5+5i} \end{align}\tag{7} $$ Since the solution in $(7)$ is given mod $(2+3i)(5+5i)=-5+25i$, adding $(1-i)(-5+25i)=20+30i$, we also get the solution $$ \begin{align} -1+10i&\equiv \phantom{-}1\pmod{2+3i}\\ -1+10i&\equiv -1\pmod{5+5i} \end{align}\tag{8} $$

Answer 2

Put $a=x+yi$ and $b=u+vi$; operating one has $(x+yi)(2+3i)+5(u+vi)(1+i)=1$ which gives $$\begin{cases}2x-3y+5u-5v=1\\3x+2y+5u+5v=0\end{cases}$$ This is a linear system of two equations with four unknowns so it should normally be infinitely many solutions. It follows, solving respect of $u,v$ $$u=\frac{1+y-5x}{10}\\v=\frac{-x-5y-1}{10}$$ Making now $1+y=10m;\space 5x=10A;\space x-4=10B$ one has $x=10n+4$ and we get a parameterization giving an infinite set of solutions in $\mathbb Z[i]$ $$\begin{cases}x=10n+4\\y=10m-1\\u=m-2-5n\\v=-(5m+n)\end{cases}$$ $\color{red}($*It is easy to verify the identity*

$[10n+4+(10m-1)i](2+3i)+[m-2-5n-(5m+n)i](5+5i)=1\color{red})$