A binary quadratic form: $nx^2-y^2=2$ For which $n\in\mathbb{N}$ are there $(x, y)\in\mathbb{N}^2$ such that $nx^2-y^2=2$ ?
 A: As pointed out by J.Claus and JavaMan, the Pell-like equation,
$$u^2-nv^2=-2\tag{1}$$
has solutions when $n$ is a prime of form $p=8m+3$ (by Legendre). Another class obviously is,
$$u^2-(m^2+2)v^2=-2$$
A third class seems to be $n = 2p$, with $p$ the same primes above, but I have no ready proof. 
A fourth is $n=3+2a+7a^2-6a^3-3a^4-4a^5+4a^6$, so,
$$(5+9a+15a^2+2a^3-4a^4-8a^5)^2-n(3+4a+4a^2)^2 = -2\tag{2}$$
Excluding $a=0,1$, I believe the parameterization above gives the fundamental solutions of $(2)$. There are other polynomial solutions to Pell equations in my site (in fact $(2)$ is a member of a 2-parameter family), and the transformations given in Wikipedia are from this site as well.
A: This looks very similar to Pell's Equation and luckily it's already been studied, such as here.
Go through their listed transformations to arrive at $$ \left(y^2 + 1\right)^2 - n(xy)^2 = 1 $$ I'm not sure how to proceed from here, as Pell's Equation doesn't have a closed form for the fundamental solution and so it can't easily be said if one of the the solutions will be in the form $ y^2 + 1 $. 
