, 2008

Solution:
We put $A(n) = 2^{n} + 3^{n}$. From Fermat's little theorem, we have $2^{p-1} \equiv 1 \pmod{p}$ and $3^{p-1} \equiv 1 \pmod{p}$, from which we conclude $A(n) \equiv 2 \pmod{p}$. Therefore, after $p-1$ steps at most, we will have repetition of the power. It means that in order to determine the minimal prime number $p$ we seek, it is enough to determine a complete set of remainders $S(p) = \{0, 1, \ldots, p-1\}$ such that $2^{n} + 3^{n} \not\equiv 0 \pmod{p}$, for every $n \in S(p)$.

For $p = 5$ and $n = 1$ we have $A(1) \equiv 0 \pmod{5}$.

For $p = 7$ and $n = 3$ we have $A(3) \equiv 0 \pmod{7}$.

For $p = 11$ and $n = 5$ we have $A(5) \equiv 0 \pmod{11}$.

For $p = 13$ and $n = 2$ we have $A(2) \equiv 0 \pmod{13}$.

For $p = 17$ and $n = 8$ we have $A(8) \equiv 0 \pmod{17}$.

For $p = 19$ we have $A(n) \not\equiv 0 \pmod{19}$, for all $n \in S(19)$.

Hence the minimal value of $p$ is $19$.