5 is quadratic combination of 8 and 3

Question

How can i proof that the only natural solutions to $8a^2-3b^2=5$ are $(a,b)=(1,1)$ and $(a,b)=(2,3)$?

I know that solutions to $8x-3y=5$ are $(x,y)\in\{(3n+1,8n+1)\}_{n\in\mathbb{N}}$, but i can't use that for my original problem.

Answer 1

Actually, there are infinitely many solutions to $8 a^2 - 3 b^2 = 5.$ Give me a minute to find the matrix.

Given a solution $(a,b)$ we get an infinite sequence of solutions by applying $$ (a,b) \mapsto (5a+3b, 8a+5b). $$

You should check it yourself, what is $$ 8 (5a+3b)^2 - 3 (8a+5b)^2 \; \; ? $$

Meanwhile, here is a portion of the Conway Topograph for $8 x^2 - 3 y^2.$ This gives enough information to find all the solutions to $8x^2 - 3 y^2 = 5.$ Put briefly, all solutions are generated by $(1,1)$ and $(2,3)$ under the mapping $ (a,b) \mapsto (5a+3b, 8a+5b) $ and its inverse $ (a,b) \mapsto (5a-3b, -8a+5b). $ For example, using the inverse takes $ (2,3) \mapsto (1, -1). $