Find $\gcd(15-15i,7-i)$ in $\mathbb Z[i]$
I have a trouble with this question because depending on how I factorizes $15-15i$ I get other sollutions:
$$7-i=(1-i)(1+2i)(2-i)$$ 1) $$15-15i=15(1-i)=3\cdot5\cdot(1-i)=3\cdot(2-i)(2+i)(1-i)$$ Then: $ \gcd(15-15i,7-i)=(1-i)(2-i)=1-3i$
2) $$15-15i=15(1-i)=3\cdot5\cdot(1-i)=3\cdot(1-2i)(1+2i)(1-i)$$ Then: $\gcd(15-15i,7-1)=(1-i)(1+2i)=3+i$
Where is the mistake?
You did not make any mistakes.
The GCD of two elements of $\mathbb{Z}[i]$ is only unique up to multiplication by a unit. The units of $\mathbb{Z}[i]$ are $\pm1$, $\pm i$.
Hence for any $a,b\in\mathbb{Z}[i]$, if $d$ is a GCD of $a$, $b$, then so is $-d$, $i\cdot d$ and $-i\cdot d$.
Since $1-3i$ is a GCD of $7-i$ and $15-15i$, $$3-i=-i\cdot(1-3i)$$
is also a GCD of $7-i$ and $15-15i$.