Actually generates
Indeed and (and the signs do make sense) thus
You might be interested in reading here.
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:
so
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.
Thanks!
-Dan
I will look at the document. Thank you!
I'm still having a problem with the derivation of (a, c) = (2, 2) from (x, y) = (3, 1). Keeping track of the signs was confusing at first, so I did the problem backward. We need the equations:
Now, the solution for c from the first equation is what gives the (+/-) in the second equation. When I plug that through the first solution that only gives me two possibilities for (x, y) and neither solve the Pell equation. In order to generate a valid solution to the Pell equation, that is (a, c) = (2, 2), we need to assume that the (+/-) in the second equation is an extra (+/-) not correlated to the (+/-) on the solution of c. I can't see how this can be.
-Dan
Actually, is comes from:
and ...
Don't forget that if is a solution to , then so is (with all 4 possibilities)
I prefer going though linear equations though:
You have (look back at your equations) : and (you use the signs that are that the same level )
Thus:
and : only 2 possibilities (either plus in both, or minus in both).
And then .
With we get:
and
Pick the - sign then : and thus
Pick the + sign then: and hence
Both and are solutions to the original equation.
Here with the linear equation we don't need to consider since the different solutions come from .