Extra Solutions of a Quadratic Diophantine Equation
Okay, I've been making a brief foray into quadratic Diophantine equations. Here's my practice equation and it's solution to Pell form. (My questions will come after the derivation.)
Let b = 2a + 1
Now take this a modulo 7:
Thus, since 7 is prime we obtain:
where y is any integer.
Plugging this in for b:
Let giving (finally!)
This is a Pell equation. One solution to this is (x, y) = (1, 0), which generates the solutions (a, c) = (0, 1); (-1, 1).
A second solution (x, y) = (-1, 0) generates (a, c) = (0, -1); (-1, -1). So far so good.
Another solution to the Pell equation is (x, y) = (3, 1), which generates the solutions (a, c) = (4, 2); (5, 4). We can generate (a, c) = (4, -2); (5, -4) from (x, y) = (-3, 1). ((x, y) = (3, -1) and (-3, -1) simply copy these solutions.)
My first question: The "solutions" (a, c) = (4, 2); (4, -2) do not solve the original equation. Why do these solutions appear?
My second question: Two solutions (a, c) = (2, -2); (2, 2) do not appear using this method. Yet I understood that all solutions to the original equation derive from solutions to the Pell equation? I can't find a solution to the Pell equation that generates them.