(Italy TST 2003) Find all strictly positive integers $(a, b, p)$ with $p$ prime such that $2^{a}+p^{b}=19^{a}$.
We rewrite the equation in the form $19^{a}-2^{a}=p^{b}$. Since 17 divides the left-hand side, we must have $p=17$. According to Zsigmondy's theorem, if $a \geq 2$, there exists a prime number dividing $19^{a}-2^{a}$ but not $19^{1}-2^{1}=17$, which is absurd. Therefore, the only solution is $(a, b, p)=(1,1,17)$.