Proving that the Diophantine equation $(11a + 5b)^2 - 223b^2 = \pm 11$ has no solutions I am working on an algebraic number theory exercise, which is to prove that $\mathbb Z[\sqrt{223}]$ has three ideal classes. I've run against the following two (really four) Diophantine equations:
$$
(11a + 5b)^2 - 223b^2 = \pm 11
$$
$$
(3a + b)^2 - 223b^2 = \pm 3
$$
I think I should be able to prove that neither of these pairs of equations has any solutions in $\mathbb Z^2$ - I have run a program to check all small values of $a$ and $b$ (less than 10,000) and found nothing, but I know that the minimum solutions to equations like this can be quite large.
What I've tried doing so far is reducing the first equation mod $11$ and mod $5$, both of which seem to give tautologies, and reducing the second equation mod $3$, which also wasn't useful. I don't know much in this area, so I'm not sure how to begin attacking the problem.
 A: There are techniques due to Dirichlet that achieve what you want in a finite number of steps. In the present case, the following ad-hoc computations do the trick.
First observe that $\alpha = 14 + \sqrt{223}$ has norm $-27$
(this implies that your second equation has a rational solution, which in turn suggests that you cannot prove it impossible by working modulo integers). Thus if there is an element of nurm $\pm 3$, one of the elements $\alpha$, $\varepsilon \alpha$ or $\varepsilon^2\alpha$ must be a cube, where $\varepsilon = 224 + 15 \sqrt{223}$ is the fundamental unit (which can be computed from the element $\beta = 15 + \sqrt{223}$ with norm $2$ via $\varepsilon = \beta^2/2$). Now you check that none of these elements is a cube.
For showing that $\alpha$ is not a cube assume that $\alpha = \gamma^3$ and $\alpha' = {\gamma'}^3$. Then $\gamma \approx 3.07$ and $\gamma' \approx -0,977$, and since $\gamma + \gamma'$ is not an integer, this is impossible.
The ideals of norm $11$ do not contribute to the class group since $16 \pm \sqrt{223}$ have norm $33$.
