Finding all integer solutions of a quadratic equation

Question

I am trying to find all integer solutions for the equation $x^2+xy+5=y^2$. I believe it is possible to apply some sort of substitution and completing the square to rewrite the equation into the form $x^2-5y^2= \pm 4$ which is a solvable Pell's equation. I can rewrite the equation as $x^2+xy-y^2=-5$. However I do not see the correct way to proceed.

Answer 1

Multiplying your equation by $4$ makes completing the square easier: $$(2x+y)^2-5y^2=-20.$$ So you need to solve $$z^2-5y^2=-20$$ with $z$ and $y$ of the same parity (but this is automatic). This means that $z=5u$ and $$y^2-5u^2=4.$$ I hope this is Pell enough for you...

Answer 2

There is a pictorial method by Conway that I like, especially for indefinite $Ax^2 + B xy + C y^2 = D$ when $B \neq 0.$ Starting with the solution $$ \left( \begin{array}{c} 1 \\ 3 \end{array} \right) $$

we find all representations of $-5$ as $$ \left( \begin{array}{cc} 1 & 1\\ 1 & 2 \end{array} \right) \left( \begin{array}{c} 1 \\ 3 \end{array} \right) $$ where $n$ is an integer that may be positive, negative, or $0$ (the identity matrix)

Put another way, we have $x_0 = -1, \; x_1 = 1, \; $ then $y_0 = 2, y_1 = 3,$ then $$ x_{n+2} = 3 x_{n+1} - x_n, $$ $$ y_{n+2} = 3 y_{n+1} - y_n \; . $$ We can go backwards to negative $n$ because $$ x_{n} = 3 x_{n+1} - x_{n+2}, $$ $$ y_{n} = 3 y_{n+1} - y_{n+2} \; . $$ The results are Lucas numbers. The first few Lucas numbers are $$ 2, 1, 3, 4, 7, 11, 18, 29, 47, 76, 123, 199, 322, 521, 843, 1364, 2207, 3571, 5778, 9349, \ldots $$