Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2015/prelims/mathematics-iv/6/b/solution

Apply the Euclidean algorithm to and :
Back-substitution gives the Bézout's identity
Thus is one solution of the Diophantine equation. Subtracting this solution from any other gives
Since and are coprime, divides . To justify that step directly, multiply by : both terms on the right are divisible by , because . Write ; substitution then gives . Therefore the complete solution set is
Every such pair satisfies the original Diophantine equation, since the terms cancel.

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