Prove n+6 in S:
n%6=1:
let a=n-2, b=n-4, We know a,b in S
n is not divisble by 2, thus
gcd(n-2,2)=gcd(n,4)=1
gcd(n-2,n-n+2)=gcd(n,n-n+4)=1
gcd(a,n-a)=gcd(n,n-b)=1
gcd(a,n)=gcd(b,n)=1
thus n is in S
n%6=2:
let a=n-1, b=n-3. We know a,b in S
We know n can't be divisble by 3, also GCD of any number and 1=1
gcd(n,1)=gcd(n,3)=1
gcd(n,n-n+1)=gcd(n,n-n+3)=1
gcd(n,n-1)=gcd(n,n-3)=1
gcd(n,a)=gcd(n,n-3)=1