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.

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