Show that the GCD of 3+4i and 7-i in Z[i] is 1

Question

I have tried proceeding by starting dividing 3+4i by 7-i and the other way as well, but don't know how to get to 1

Answer 1

$$ (3+4i)(-1+i)=-7-i $$ $$ (7-i)+(-7-i)=-2i $$ If $\beta=\gcd(3+4i,7-i)$, then $\beta|(-7-i)$, and so $\beta|(-2i)$.

Now notice that the function $N:\mathbb{Z}[i]\to\mathbb{N}$ defined by $N(a+bi)=a^2+b^2$ is a multiplicative function. Then we must have $N(\beta)|N(3+4i)$, $N(\beta)|N(7-i)$ and $N(\beta)|N(-2i)$. The latter gives $N(\beta)=1$, $N(\beta)=2$ or $N(\beta)=4$ and since $2\nmid 25=N(3+4i)$, it must be that $N(\beta) = 1$. Then it is straightforward to see that $\beta$ is a unit.

Answer 2

Let $\alpha$ and $\beta$ be two Gaussian integers. Consider the parallelogram $P$ defined by $0$, $\alpha$, $\beta$, $\alpha+\beta$. When you multiply $\alpha$ and $\beta$ by $\delta$, you get another parallelogram, whose area is $N(\delta)=\delta\bar\delta$ times the area of $P$.

For $\alpha=3+4i$ and $\beta=7-i$, the area of $P$ is $31$, which is prime.

Therefore, any common divisor must have norm $1$, and so is a unit.