1. ## 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.)
$a^2 + a + 2 = 2c^2$

$a^2 + a + \frac{1}{4} - \frac{1}{4} + 2 = 2c^2$

$\left ( a + \frac{1}{2} \right ) ^2 + \frac{7}{4} = 2c^2$

$(2a + 1)^2 + 7 = 8c^2$

Let b = 2a + 1

$b^2 + 7 = 8c^2$

Now take this a modulo 7:
$b^2 \equiv c^2~\text{(mod 7)}$

Thus, since 7 is prime we obtain:
$b \equiv \pm c$

so
$b = 7y \pm c$

where y is any integer.

Plugging this in for b:
$(7y \pm c)^2 + 7 = 8c^2$

$49y^2 \pm 14yc + c^2 + 7 = 8c^2$

$49y^2 \pm 14yc - 7c^2 = -7$

$7y^2 \pm 2yc - c^2 = -1$

$c^2 \mp 2yc + y^2 - y^2 - 7y^2 = 1$

$(c \mp y)^2 - 8y^2 = 1$

Let $x = c \mp y$ giving (finally!)
$x^2 - 8y^2 = 1$

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

2. Actually $(x,y)=(3,1)$ generates $(a,c)=(2,2)$

Indeed $c = x - y = 2$ and $b = 7\cdot y - c = 5$ (and the signs do make sense) thus $a = \frac{b-1}{2} = 2$

You might be interested in reading here.

3. Originally Posted by PaulRS
Actually $(x,y)=(3,1)$ generates $(a,c)=(2,2)$

Indeed $c = x - y = 2$ and $b = 7\cdot y - c = 5$ (and the signs do make sense) thus $a = \frac{b-1}{2} = 2$

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:

$b^2 - 8c^2 = -7$

$b = 7y \pm c$

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

4. Actually, $(a,c)=(2,2)$ is comes from:

$b^2-8c^2 = -7$ and $b=7\cdot 1 - c$ ...

Don't forget that if $(b,c)$ is a solution to $b^2-8c^2=-7$, then so is $(\pm b,\pm c)$ (with all 4 possibilities)

I prefer going though linear equations though:

You have (look back at your equations) : $x = c \mp y$ and $b = 7\cdot y \pm c$ (you use the signs that are that the same level )

Thus:
$c = x \pm y$ and $b = 7\cdot y \pm c$ : only 2 possibilities (either plus in both, or minus in both).

And then $a = \frac{b-1}{2}$.

With $(x,y)=(3,1)$ we get:

$c = 3 \pm 1$ and $b = 7\cdot 1 \pm c$

Pick the - sign then : $c = 2$ and $b = 5$ thus $a = 2$

Pick the + sign then: $c = 4$ and $b = 7\cdot 1 + 4 = 11$ hence $a = 5$

Both $(a,c)=(2,2)$ and $(a,c)=(5,4)$ are solutions to the original equation.

Here with the linear equation we don't need to consider $(\pm b,\pm c)$ since the different solutions come from $(\pm x,\pm y)$.