= Solution
The <Euclidean algorithm> gives
$$
100=79+21,\qquad79=3\cdot21+16,\qquad21=16+5,\qquad16=3\cdot5+1.
$$
Back substitution produces the <Bézout's identity>
$$
1=16-3(21-16)=4(79-3\cdot21)-3\cdot21
=19\cdot79-15\cdot100.
$$
Hence 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 $k\geq0$.
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