1. Find all natural numbers $m$ and $n$ that satisfy the equation $\frac{3}{m}+\frac{5}{n}=1$.
$\mathbf{I} / 1$. We rewrite the equation $\frac{3}{m}+\frac{5}{n}=1$ as $5 m+3 n=m n$ or $(m-3)(n-5)=15$. Since $15=1 \cdot 15=3 \cdot 5$ and $m-3$ and $n-5$ are integers, the possible pairs $(m-3, n-5)$ are $(-1,-15),(-3,-5),(-5,-3),(-15,-1),(1,15),(3,5),(5,3)$ and $(15,1)$. In the first four cases, the corresponding $m$ and $n$ are not both positive, while the other four cases give us the solutions $(4,20),(6,10),(8,8)$ and $(18,6)$.

Writing the equation $(m-3)(n-5)=15$ or an equivalent form that clearly shows there are only finitely many solutions: 3 points. Complete analysis of cases and all 4 solutions: 4 points.

If the contestant does not analyze all cases, award 1 point for every 2 cases analyzed correctly. If the contestant only lists all correct solutions without proving that these are all: 2 points.