 Solution

Suppose there were only finitely many prime numbers congruent to $2$ modulo $3$, and list them as $p_1,\dots ,p_k$. ConsiderNo $p_i$ divides $N$, and $3$ does not divide $N$. In the prime factorization of $N$, not every prime factor can be congruent to $1$ modulo $3$, since their product would then also be congruent to $1$, whereasThus some prime factor is congruent to $2$ modulo $3$, contradicting the completeness of the list. Hence

Now let $p$ be prime. If $p\ne 3$, then either $p=2$, givingor $p$ is not divisible by $3$, so $p^2\equiv 1(\mathrm{mod}3)$. In the latter caseand is greater than $3$, hence is composite. For $p=3$ the number is $19$, which is prime. Therefore