Prove that there are infinitely many $n \in \mathbb{N}$ such that $6n - 1, 6n + 1$ are composite Let m be a integer. Then if $6n+1$ is a composite number we have that $\operatorname{lcd}(6n+1,m)$ is not just $1$, because then $6n+1$ would be prime. Also this is for $6n-1$. Now I must find all $m$ that this is correct for. I don't know if I'm gonna do it with induction by $n$ or by $m$.
Or is this wrong? Can someone help me please?
 A: Consider the numbers
$$6(35k+1)-1  \quad\text{and}  \quad 6(35k+1)+1$$
for $k=1,2,3,\dots$. The first is divisible by $5$ and the second is divisible by $7$ for all $k$.
A: If not, then either $6n-1$ or $6n+1$ is prime for all $n\geq N$ for some large $N$.
Then at least every $6$th number is prime for sufficiently large $N$, and then $\pi(x)\geq \frac{1}{6}x$, where $\pi(x)$ is the prime counting function. This contradicts the prime number theorem.
A: HINT: Suppose $x$ is divisible by $5, 6,$ and $7$ - what can you say about $(x+6)-1$ and $(x+6)+1$?
A: I like the other answers, but here is one more. Let $n=6k^2$ where $k$ is any positive integer whose decimal representation ends in $2$ or in $8$. Then 
$6\cdot6k^2-1=(6k)^2-1=(6k-1)(6k+1)$, composite. 
$k^2$ ends in $4$, hence $6\cdot6k^2$ ends in $4$ too, and $6\cdot6k^2+1$ ends in $5$, it is clearly divisible by $5$ (and by something else, using that $36k^2+1>5$). 
A: Choose a prime $p>3$ then $\gcd(6,p) = 1$. Then there are $x,y$ such that
$6x+py = 1$.
Furthermore, $6(x+np) -1 = p(6n-y)$ for all $n$, so by choosing $n$
appropriately
we can find $x_1,y_1 $ with $x_1 >p $ such that $6x_1 -1 = p y_1$ and
hence $6x_1 -1$ is composite.
Similarly, since $6(np-x) +1 = p(y+6n)$ for all $n$, hence we can find
$x_2,y_2$ with $x_2 > p$ such that $6x_2+1 = p y_2$ and hence $6x_2 +1$ is composite.
To repeat, choose a new prime $p' > \max(6x_1-1, 6 x_2+1)$.
