Show, that for every k$\in \mathbb N$ , $2^n+3^n-1,2^n+3^n-2,...,2^n+3^n-k$ are all composite for some $n$

Question

Show that for every $k\in \mathbb N$ there exists a number $n\in\mathbb N$ ,such that

$2^n+3^n-1,2^n+3^n-2,...,2^n+3^n-k$

are all composite.

Answer 1

Let $k\in\mathbb N$ and let $N>0$ be a sufficiently large positive integer. Let $p_1,\ldots,p_k$ be primes dividing $2^N+3^N-1,\ldots,2^N+3^N-k$ respectively and let $q=(p_1-1)\cdots(p_k-1)$.

Then for any $m\in\{1,\ldots,k\}$, $p_m\mid2^N+3^N-m\equiv2^{N+q}+3^{N+q}-m\pmod{p_m}$.
At least, if $p_m\neq3$, which means we'll have to make sure the $p_m$'s can be chosen to be different from $3$. Indeed: $3^N<2^N+3^N-m<3^{N+1}$ as soon as $2^N>m$, so we just have to take $N>\log_2(k)$.

With this choice of $p_m$'s, every $2^{N+q}+3^{N+q}-m$ is composite and we can take $n=N+q$.

Note that this gives an explicit upper-bound for the least such $n$, namely $$n\leq\lfloor1+\log_2k\rfloor+\prod_{m=1}^k(2^{\lfloor1+\log_2k\rfloor}+3^{\lfloor1+\log_2k\rfloor}-m-1).$$ (This rougly grows like $k^{k\log_2(3)}\approx(k^k)^{1.585}$.)