Let $n>2$ be a natural number.
Prove that there exist two different prime numbers $p,q<\frac{n}{2}$ which don't divide $n$ such that $p\not\equiv q<6>.$
My attempt:
if $n$ is divisible by $2$ or $3$ we get the result.
I try to prove the following:
there exists a natural number $m<\frac{n}{2}$ and $gcd(m,n)=1$ such $m\equiv 5<6>$ for every $n$ big enough,
because I hope that this will help me to solve the problem.
I hope you will help me too.
Regards.
This is more of a rationale than a hard and fast proof. First note that per the comment of lulu, $n>2$ is an inadequate lower limit. In fact, for $n\le 6$, there are not even two primes smaller than $n/2$. Let's assume there is some reasonable value for $n$ that will satisfy the other constraints, without specifically identifying it here.
If $2\not \mid n,\ 3\not \mid n$, then $p=2,\ q=3$ is a solution.
If $2\mid n,\ 3\not \mid n$, then $p=3$. In order that there be no possible $q$, $n$ must contain as a factor every prime $3<r_i\le \frac{n}{2}$. For reasonable sized $n$, $2\prod r_i >n$. So this is not a possibility, as there will be some $r_j=q$.
Similarly, if $3\mid n,\ 2\not \mid n$, then $p=2$. In order that there be no possible $q$, $n$ must again contain as a factor every prime $3<r_i\le \frac{n}{2}$. For reasonable sized $n$, $3\prod r_i >n$. So this is not a possibility, as there will be some $r_j=q$.
So the remaining case is $6\mid n$. Here, there will be a suitable $p,q$ unless $n$ contains as a factor every prime of the form $6k-1$ OR every prime of the form $6k+1$ (or both). Let's look at these cases empirically.
$5$ is the smallest prime of the form $6k-1$, and $6\cdot 5=30$, For $n=30,\ \frac{n}{2}=15$, we have the additional prime $11$ of the form $6k-1$. If we include that in the product, then $n=330,\ \frac{n}{2}=115$ and our problem becomes worse in that there are many more primes of the form $6k-1$ between $11$ and $115$.
$7$ is the smallest prime of the form $6k+1$, and $6\cdot 7=35$, For $n=35,\ \frac{n}{2}=17.5$, we have the additional prime $13$ of the form $6k+1$. If we include that in the product, then $n=455,\ \frac{n}{2}=227.5$ and our problem again becomes similarly worse.
The statements here give a reasonable basis to credit the truth of your proposition, but aren't quite tight enough to constitute a formal proof.