$ k^2+2n, k^2, k^2+1, 2n, 2n+1$, (for some $n$) if $k$ is even and $0 < n < k$. I have no idea of how to prove that. I'm working on Legendre's conjecture.
Update 1: Yes, for $n=0$ all $l$ divides $2n$, but I need to work on some non-constructive proof over $n$, because on another step I also need to prove there's some $n$ such that $k^2+1, 2n$ are coprimes and $k^2,2n+1$ are coprimes ($0$ is coprime with 1, but $k^2+1 \neq 1$). I guess $n=0$ must be an exceptional case.
Update 2: $0 < n < k \implies k^2+2n+1 < (k+1)^2$, therefore $\lfloor \sqrt{k^2+2n+1} \rfloor = k$. If we want to prove every $2 \leq l \leq k$ divides at least one of the numbers above (Line 1) for some $n$, we can prove that $[(k^2+2n)(k^2)(k^2+1)(2n)(2n+1)]^k \equiv 0 \mod k!$, for some $0 < n < k$. But it's also too complicated.