It means : Prove that "Have infinity primes $6k+5$."

You can suppose have limited primes numbers form $6k+5$

Indeed! This is true by Dirichlet's theorem, since gcd(6,5)=1

Consider $N=6p_{1}...p_{n}-1$

Another way is considering prime divisors of $n^2-n+1$, not equal to 3. These primes are $p\equiv 5\pmod 6$

@above: I think prime divisors of $n^2-n+1$ not equal to $3$ will generally be congruent to $1$ modulo $6,$ not $5$ modulo $6.$

Taking all natural numbers as $n=6k+r$ with $r=0, 1,..., 5$ and later removing all multiples of $2$ and $3$, we obtain that all the prime numbers are of the form $6k+1$ and $6k+5$. The result follows easily after this.

Key concept : Product of any two numbers of the form $6k+1$ is again of the form $6j+1$.

Each term is of the form $6k+5$, but since $\gcd(5,6)=1,$ Dirichlet's Theorem on $(a,b)=(6,5)$ finishes. $\blacksquare$

Taco12 wrote: Each term is of the form $6k+5$, but since $\gcd(5,6)=1,$ Dirichlet's Theorem on $(a,b)=(6,5)$ finishes. $\blacksquare$ I think the goal of the exercise is to do it without Dirichlet's Theorem. Claim: Infinitely many primes $p\equiv 5\mod 6$. Proof: Note that all primes $p\geq 5$ are of the forms $1\mod 6$ or $5\mod 6$. Assume there are only $n\in\mathbb{N}$ of them and call them $\{p_i\}_{i=1}^n$. If $n$ is even, let $P=p_1\cdots p_n+4$. If $n$ is odd, let $P=p_1\cdots p_n+6$. Notice, $P\equiv 5\mod 6$ in both cases and so because the product of $1\mod 6$ primes is $1\mod 6$, the prime factorization of $P$ must have a prime factor $p\equiv 5\mod 6$ But note that $\forall i(p_i\not| P)$. Therefore $p\notin\{p_i\}_{i=1}^n$. This is a contradiction since it shows $\{p_i\}_{i=1}^n$ is not a complete list of $5\mod 6$ primes. $\blacksquare$