I came across a problem in Dudley's Elementary Number Theory. Part (a) is to find one $n$ such that both $6n+1$ and $6n-1$ are composite. So I stumbled across $n=50$, which gives $299=13 \cdot 23$, and $301=7 \cdot 43$
The motivation for this was knowing that $9|10^k -1$ and $7|10^{2k}+1$ and trying to find an even power of ten (obviously $300$ is not a power of ten).
Part (b) asks to prove that there are infinitely many $n$ such that both are composite.
How do we show this?
Setting $6n \pm 1 = 10^{2k} \pm 1$ didn't work.
I looked at $(6n+1)(6n-1) = 36n^2-1 = 6(6n^2) -1$, so I can generate composite numbers of the form $6m -1$, but what about the corresponding $6m+1$ ?
Should I be using my result from (a) to generate these pairs?
I looked at the solutions to this related question, but they seemed focused on proving that both are prime.
Other methods that I've considered are using modular arithmetic (though that hasn't been covered at this point in the book) and using the well-ordering principle in some way…
(Hints are acceptable, as I want to be an autonomous mathematician, as much as possible)
Added
If we let $n = 6^{2k} = 36^k$, then $6n \pm 1 = 6^{2k+1} \pm 1$, which (for $k \geq 1$) factors into $(6 \pm 1)(6^{2k} \mp … +1)$
Is this correct? Here I am using a factorization of sums/differences of odd powers, which "hasn't been covered" (which isn't itself a problem).
Best Answer
Hint: Instead of even powers of $10$, try odd powers of $6$.