Past exam of the mathematics course of the University of Oxford / 2015 / Prelims / Mathematics IV / 6 / b

Apply the Euclidean algorithm to $19$ and $11$:Back-substitution gives the Bézout's identityThus $(x,y)=(-4,7)$ is one solution of the Diophantine equation. Subtracting this solution from any other givesSince $19$ and $11$ are coprime, $11$ divides $x+4$. To justify that step directly, multiply $1=-4\cdot 19+7\cdot 11$ by $x+4$: both terms on the right are divisible by $11$, because $19(x+4)=-11(y-7)$. Write $x+4=11t$; substitution then gives $y-7=-19t$. Therefore the complete solution set isEvery such pair satisfies the original Diophantine equation, since the $t$ terms cancel.