doesn't exist an $N$ s.t. all $n \ge N$ satisfy an equation.

Question

I came across this problem on my own and i'm asking for any potential techniques/strategies/hints for attacking it.

Prove that there does not exist an $N$ such that for every natural number $n \ge N$; $n = {cx \pm 1\over 6}$ for some odd $c \ge 3$, for some odd $x \ge 5$

Thanks in advance.

Answer 1

I believe this problem is equivalent to the Twin Primes conjecture (TPC)

To see that TPC implies your question: Suppose we had such an $N$. Choose Twin primes larger than $6N$. Then we have the triple $\{m-1,m,m+1\}$ where $m\pm 1$ are both prime. Clearly $6$ divides $m$ and $n=\frac m6$ is a counterexample.

To see that your question implies the TPC...suppose the TPC is false. Then choose $N$ so big that there are no Twin Primes greater than $N$. Then for $n≥N$ we know that at least one of $6n\pm 1$ is composite and thus we can find $c,x$ as desired.