Quadratic Diophantine equation - Find all integer solutions

Question

I have a Quadratic Diophantine equation.

I am sorry I didn't get how to show correctly. So I used a picture

How can I find all integer solutions? Should I do something with Z(14)?

Answer 1

The integer equation $11 x^2 - 14 y^2 = 1$ is impossible. However, I do not see that a proof is available for a beginner. There is no simple proof with congruences. There is one using, in effect, continued fractions.

The principal genus of this discriminant is two forms, $$ x^2 - 154 y^2 $$ $$ 2 x^2 - 77 y^2, $$ and your form $11x^2 - 14 y^2$ is $SL_2 \mathbb Z$ equivalent to $2x^2 - 77 y^2$ and simply does not integrally represent $1.$

The other genus is their negatives, $$ -x^2 + 154 y^2 $$ $$ -2 x^2 + 77 y^2, $$

Here are a list of the class group, Gauss-Lagrange reduced forms, and the Lagrange cycle of your form

    1.             1          24         -10   cycle length            10
    2.            -1          24          10   cycle length            10
    3.             2          24          -5   cycle length             8
    4.            -2          24           5   cycle length             8

  form class number is   4

jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./indefCycle 11 0 -14

  0  form             11           0         -14  delta      0
  1  form            -14           0          11  delta      1
  2  form             11          22          -3


          -1          -1
           0          -1

To Return  
          -1           1
           0          -1

0  form   11 22 -3   delta  -7     ambiguous  
1  form   -3 20 18   delta  1
2  form   18 16 -5   delta  -4
3  form   -5 24 2   delta  12
4  form   2 24 -5   delta  -4     ambiguous  
5  form   -5 16 18   delta  1
6  form   18 20 -3   delta  -7
7  form   -3 22 11   delta  2
8  form   11 22 -3


  form   11 x^2  + 22 x y  -3 y^2 

minimum was   2rep   x = 5   y = 39 disc 616 dSqrt 24  M_Ratio  4.760331
Automorph, written on right of Gram matrix:  
2419  5148
18876  40171
=========================================
jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ 

Answer 2

Second answer: your problem is equivalent to showing, with integers $x,y,$ that $$ x^2 - 154 y^2 \neq 2 $$ This may be more familiar, it is a matter of constructing the continued fraction for $\sqrt {154}$ and showing that, for any "convergent" $\frac{p}{q},$ we always have $p^2 - 154 q^2 \neq 2.$

Method described by Prof. Lubin at Continued fraction of $\sqrt{67} - 4$

$$ \sqrt { 154} = 12 + \frac{ \sqrt {154} - 12 }{ 1 } $$ $$ \frac{ 1 }{ \sqrt {154} - 12 } = \frac{ \sqrt {154} + 12 }{10 } = 2 + \frac{ \sqrt {154} - 8 }{10 } $$ $$ \frac{ 10 }{ \sqrt {154} - 8 } = \frac{ \sqrt {154} + 8 }{9 } = 2 + \frac{ \sqrt {154} - 10 }{9 } $$ $$ \frac{ 9 }{ \sqrt {154} - 10 } = \frac{ \sqrt {154} + 10 }{6 } = 3 + \frac{ \sqrt {154} - 8 }{6 } $$ $$ \frac{ 6 }{ \sqrt {154} - 8 } = \frac{ \sqrt {154} + 8 }{15 } = 1 + \frac{ \sqrt {154} - 7 }{15 } $$ $$ \frac{ 15 }{ \sqrt {154} - 7 } = \frac{ \sqrt {154} + 7 }{7 } = 2 + \frac{ \sqrt {154} - 7 }{7 } $$ $$ \frac{ 7 }{ \sqrt {154} - 7 } = \frac{ \sqrt {154} + 7 }{15 } = 1 + \frac{ \sqrt {154} - 8 }{15 } $$ $$ \frac{ 15 }{ \sqrt {154} - 8 } = \frac{ \sqrt {154} + 8 }{6 } = 3 + \frac{ \sqrt {154} - 10 }{6 } $$ $$ \frac{ 6 }{ \sqrt {154} - 10 } = \frac{ \sqrt {154} + 10 }{9 } = 2 + \frac{ \sqrt {154} - 8 }{9 } $$ $$ \frac{ 9 }{ \sqrt {154} - 8 } = \frac{ \sqrt {154} + 8 }{10 } = 2 + \frac{ \sqrt {154} - 12 }{10 } $$ $$ \frac{ 10 }{ \sqrt {154} - 12 } = \frac{ \sqrt {154} + 12 }{1 } = 24 + \frac{ \sqrt {154} - 12 }{1 } $$

Simple continued fraction tableau:
$$ \begin{array}{cccccccccccccccccccccccccc} & & 12 & & 2 & & 2 & & 3 & & 1 & & 2 & & 1 & & 3 & & 2 & & 2 & & 24 & \\ \\ \frac{ 0 }{ 1 } & \frac{ 1 }{ 0 } & & \frac{ 12 }{ 1 } & & \frac{ 25 }{ 2 } & & \frac{ 62 }{ 5 } & & \frac{ 211 }{ 17 } & & \frac{ 273 }{ 22 } & & \frac{ 757 }{ 61 } & & \frac{ 1030 }{ 83 } & & \frac{ 3847 }{ 310 } & & \frac{ 8724 }{ 703 } & & \frac{ 21295 }{ 1716 } \\ \\ & 1 & & -10 & & 9 & & -6 & & 15 & & -7 & & 15 & & -6 & & 9 & & -10 & & 1 \end{array} $$

$$ \begin{array}{cccc} \frac{ 1 }{ 0 } & 1^2 - 154 \cdot 0^2 = 1 & \mbox{digit} & 12 \\ \frac{ 12 }{ 1 } & 12^2 - 154 \cdot 1^2 = -10 & \mbox{digit} & 2 \\ \frac{ 25 }{ 2 } & 25^2 - 154 \cdot 2^2 = 9 & \mbox{digit} & 2 \\ \frac{ 62 }{ 5 } & 62^2 - 154 \cdot 5^2 = -6 & \mbox{digit} & 3 \\ \frac{ 211 }{ 17 } & 211^2 - 154 \cdot 17^2 = 15 & \mbox{digit} & 1 \\ \frac{ 273 }{ 22 } & 273^2 - 154 \cdot 22^2 = -7 & \mbox{digit} & 2 \\ \frac{ 757 }{ 61 } & 757^2 - 154 \cdot 61^2 = 15 & \mbox{digit} & 1 \\ \frac{ 1030 }{ 83 } & 1030^2 - 154 \cdot 83^2 = -6 & \mbox{digit} & 3 \\ \frac{ 3847 }{ 310 } & 3847^2 - 154 \cdot 310^2 = 9 & \mbox{digit} & 2 \\ \frac{ 8724 }{ 703 } & 8724^2 - 154 \cdot 703^2 = -10 & \mbox{digit} & 2 \\ \frac{ 21295 }{ 1716 } & 21295^2 - 154 \cdot 1716^2 = 1 & \mbox{digit} & 24 \\ \end{array} $$

Answer 3

If you are asking for $(x, y)$ pairs that solve the equation: $$11x^2-14y^2=1$$ you can use the on solver at here to show that there are no solutions in integers.