Solving Pell’s Equation for $x^2 -7y^2 = 1$ for the first three integral solutions.

elementary-number-theorynumber theory

Like the title states my goal is the find the first three integral solutions of the Diophantine equation. I know $x^2 -7y^2 = 1$ is a Pell's equation where $d = 7$. I found the minimal solution to be $(x,y) = (8,3)$ through brute force. I then found the next pair by using the equation:

$(x + \sqrt{7}y)^i(x-\sqrt{7}y)^i = 1$

$i = 1$ would give me the minimal solution $(8,3)$, so I went to $i = 2$.

$(x + \sqrt{7}y)^2(x-\sqrt{7}y)^2 = 1$

$\left(x^2 + 7y^2 + 2xy\sqrt{7}\right) \left(x^2 + 7y^2 – 2xy\sqrt{7}\right) = 1$

$(x^2 + 7y^2)^2 – 7(2xy)^2 = 1$ (This is of the form $X^2 – 7Y^2 = 1$)

I then let $X = x^2 + 7y^2$ and $Y = 2xy$, so the equation becomes the desired

$X^2 – 7Y^2 = 1$

I then plugged in my minimal solution $(8,3)$ and found $X = (8^3+7(3^3)) = 127$ and $Y = 2(8)(3) = 48$, so my new pair is $(127,48)$. To find the next solution, I let $i = 3$.

$(x + \sqrt{7}y)^3(x-\sqrt{7}y)^3 = 1$

And this is where I got stuck. I tried a few methods to get this new equation into the form of $X^2 – 7Y^2 = 1$, but have been unsuccessful. I tried expanding out the equations completely getting:

$x^6 -21x^4y^2+147x^2y^4-343y^6 = 1$

But I'm fairly sure that's not the right direction. The other method I tried was factoring out a $(x + \sqrt{7}y)^2(x-\sqrt{7}y)^2$, so I got:

$[(x + \sqrt{7}y)(x-\sqrt{7}y)](x + \sqrt{7}y)^2(x-\sqrt{7}y)^2$

which simplifying I got

$(x^2-7y^2) \left( (x^2 + 7y^2)^2 – 7(2xy)^2 \right)$

I'm really not sure how to properly factor this cubic to get to the desired function form. Any help would be greatly appreciated!

Best Answer

Consider $$(8+3\sqrt7)^2=127+48\sqrt7,$$ $$(8+3\sqrt7)^3=(8+3\sqrt7)(127+48\sqrt7)=2024+765\sqrt7,$$ $$(8+3\sqrt7)^4=(8+3\sqrt7)(2024+765\sqrt7)=32257+12192\sqrt7$$ etc. Then the first few solutions are $(8,3)$, $(127,48)$, $(2024,765)$, $(32257,12192)$ etc.

Related Solutions

How to Generate Solutions for Quadratic Diophantine Equations

The "topograph" for $x^2 - 13 y^2$ is definitely more complicated than the previous ones, because the continued fraction for $\sqrt {13}$ has period 5, your two previous examples had period 1. Confirming the "automorph" matrix, which just preserves the quadratic form:

=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=

gp-pari 
? 
? 
? form = [ 1,0; 0,-13]
%1 = 
[1 0]

[0 -13]

? 
? a = [649, 2340; 180, 649]
%2 = 
[649 2340]

[180 649]

? 
? atranspose = mattranspose(a)
%3 = 
[649 180]

[2340 649]


? 
?  atranspose  * form * a
%5 = 
[1 0]

[0 -13]

=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=

The pairs of numbers in green are vectors in the plane. Two basic properties. First, each shows its value for $x^2 - 13 y^2.$ For example, in the first occurrence of 4, we see the (column) vector $(11,3),$ and we can easily confirm that $11^2 - 13 \cdot 3^2 = 4. $ Next, around any point where three purple line segments meet (even if two are parallel), one of the three green vectors is the sum of the other two. For example, $$ (4,1) + (7,2) = (11,3). $$ As long as we just continue to the right, we can continue getting all positive entries in green.

Oh: you said you can do continued fractions. It happens that you can find all representations of 4 and 1 using the continued fraction of $\sqrt {13},$ so you can confirm a good deal of the Conway diagram, the vectors in green, whatever.

=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=

enter image description here

=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=

enter image description here

=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=

jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./indefCycle
Input three coefficients a b c for indef f(x,y)= a x^2 + b x y + c y^2 
1 0 -13

  0  form              1           0         -13  delta      0
  1  form            -13           0           1  delta      3
  2  form              1           6          -4


          -1          -3
           0          -1

To Return  
          -1           3
           0          -1

0  form   1 6 -4   delta  -1
1  form   -4 2 3   delta  1
2  form   3 4 -3   delta  -1
3  form   -3 2 4   delta  1
4  form   4 6 -1   delta  -6
5  form   -1 6 4   delta  1
6  form   4 2 -3   delta  -1
7  form   -3 4 3   delta  1
8  form   3 2 -4   delta  -1
9  form   -4 6 1   delta  6
10  form   1 6 -4

=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=

[Math] solving cubic diophantine equation

You know that $x$ and $y$ cannot both be greater than or equal to $3$, because this would give you too large of a sum. And $x$ and $y$ must both be less than $4$ by similar reasoning. So in this case, that's probably the best way to approach the problem: Just try $1,2,3$ for $x$ and $y$, where they are not both $3$. And note you can assume $x \leq y$ by symmetry, and modify your answer after the fact based on symmetry and the solutions you get.

Related Question