## Compute Bezout’s identity in the Gaussian integers

Find Gaussian integers $\xi$ and $\eta$ such that $(3+13i)\xi + (2+5i)\eta = 1$.

Using the division algorithm for Gaussian integers, we see that $3+13i = 2(2+5i) + (-1+3i)$ and $2+5i = (1-i)(-1+3i) + i$. Back substituting, and noting that $(i)(-i) = 1$, we see that $(1+i)(3+13i) - (2+3i)(2+5i) = 1$.