Thread: 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:


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

Actually generates

Indeed and (and the signs do make sense) thus

You might be interested in reading here.

Originally Posted by PaulRS

Actually generates

Indeed and (and the signs do make sense) thus

You might be interested in reading here.

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 .