How do you find solutions to $2 x^2 +3 x +1 = y^2$ using integers for $x,y$ I am trying to solve an equation in integers to give a square number.
$$2 x^2 + 3 x +1 = y^2$$
while also satisfying $x=k^2 * n$ where $n$ is a very large integer given to us and $k$ can be any integer chosen to form a solution. The $y$ can be any integer needed to form a solution.
I am looking for a method to find solutions in integers only.
I am new to this kind of equation and can't tell easy from hard from impossible.
This is not school work.  
Thanks for your help.
 A: I'm assuming you want it to be a perfect square, otherwise the problem is trivial.
Notice that your equation is $(2x+1)(x+1) = y^2$. Since $\gcd (2x+1, x+1) = \gcd (x, x+1) = 1 $, hence we need both $x+1$ and $2x+1$ to be perfect squares. 
We thus want $(2x+1) - 2(x+1) = -1$. This is Pell's equation of the form $X^2 - 2Y^2 = -1$, and has solutions $X_k + \sqrt{2}Y_k = ( 7+ 5 \sqrt{2}) ( 3 + 2 \sqrt{2})^k$. I would suggest that you look determine this list first, and hence obtain all possible values of $x$ such that $2x^2 + 3x +1$ is a perfect square. For example, we have $(X_k, Y_k) = (7, 5), (41, 29), (239, 169), \ldots$ which gives $x = 24, 840, 28560, \ldots $. [I forgot the 'trivial solution $(1, 1)$ which yields $x=0$.]
Now, we also want $x+1 - x = 1$, which is of the form $X^2 - nY^2 = 1$ (Pell's equation). While this equation always has solutions in integers, it can be extremely hard to determine the initial (non-trivial) solution. Having done that, generate the list of possible $x$, and compare it to the previous list. 
A: As a coda to the (excellent) answer of robjohn and using the notation given there, one is led to solve the system of equations
$$
2q^2-p^2= q^2-n k^2 =1.
$$
For fixed $n$, such systems always have at most finitely many solutions, via an old theorem of Siegel (they define a genus $1$ curve, indeed an elliptic curve). In fact, one can find an absolute bound upon the number of solution triples $(p,q,k)$, independent of $n$ (probably, if we restrict to positive integers, a bound of $2$ suffices), in contrast to elliptic curves in Weierstrass form. 
There are a number of methods for finding all such solutions, given $n$, ranging from elementary to less so (the standard algorithm uses lower bounds for linear forms in logarithms). Googling "simultaneous Pell equations'' is a good place to start.
A: For a small n =6 you have the solution x =24 y =35.
