Past exam of the mathematics course of the University of Cambridge / 2004 / ia / Paper 4 / 1E / a

The Euclidean algorithm givesBack substitution produces the Bézout's identityHence a positive integer solution is $\boxed{m=19,n=15}$. More generally, subtracting this identity from any other solution shows $79(m-19)=100(n-15)$. Since $79$ and $100$ are coprime, the integer solutions are $m=19+100k$, $n=15+79k$; both are positive precisely when $k\ge 0$.