If $n\geq 6$ natural number then $\exists p,q\leq n-1$ prime number such that : $(n-q,n-p)=1$

Question

Question :

Prove that for every even integer greater than $6$ there exist two prime numbers $p$ and $q$ such that : $p$ and $q$ are less than $n-1$ and $n-p$ and $n-q$ are relatively prime

I don't know how I starte ? To prove that !

Can give me some ideas to approach ?

Answer 1

Let $n$ be even and $q=3,p=5$. Then $(n-p,n-q)=(n-3,n-5)=(2,n-5)=1$ This is the same idea as in the comment by gandalf61, which came in while I was writing this.