infinitely many perfect squares in $\{2n^2+kn+l\colon n\in\mathbb{N}\}$

Question

is there a pair $(k,l)$ of natural numbers such that the set $\{2n^2+kn+l\colon n\in\mathbb{N}\}$ contains infinitely many perfect squares?

Answer 1

Use $k=4$ and $l=3$. Then $2n^2+kn+l=2(n+1)^2+1$.

It is a standard result that the Pell equation $$x^2-2y^2=1$$ has infinitely many solutions in positive integers. So there are infinitely many squares of the form $2(n+1)^2+1$.

Answer 2

$2n^2 + kn + l = m^2 \Rightarrow 2n^2+kn +l-m^2 = 0\Rightarrow \triangle = k^2-8(l-m^2) = t^2\Rightarrow l = \dfrac{k^2+8m^2 - t^2}{8}$. This easily has integer solutions in $k$, then in $l$. In fact, you start out by selecting any $k$, then choose a $t$ such that $8 \mid (k^2-t^2)$, then $l$ is an integer, and you are done. This solves one integer solution, and to obtain infinitely many integer solutions, you can choose $k, t$ as a function of a third parameter $s$ which is easy to do.