Determine the minimal prime number $p>3$ for which no natural number $n$ satisfies

$$
2^{n}+3^{n} \equiv 0(\bmod p)
$$
We put $A(n)=2^{n}+3^{n}$. From Fermat's little theorem, we have $2^{p-1} \equiv 1(\bmod p)$ and $3^{p-1} \equiv 1(\bmod p)$ from which we conclude $A(n) \equiv 2(\bmod p)$. 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 $2^{n}+3^{n} \not \equiv 0(\bmod p)$, for every $n \in S(p)$.

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

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

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

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

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

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

Hence the minimal value of $p$ is 19.