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The pell-like equation $$x^2-101y^2=-71$$ has the rational solution $(x,y)=(\frac{25}{2},\frac{3}{2})$

Can I use this rational point to find out , whether an integral solution exists ? If yes, can I also find it out for an arbitary equation $$Ax^2+Bxy+Cy^2+Dx+Ey+F=0$$ Again, we can assume that we know a rational solution.

I read somewhere that a rational point allows a parametrization, but I am not sure whether this helps to solve my problem.

Peter
  • 84,454
  • Reason for this question : The Magma-Calculator finds a rational solution of a conic, if it exists. PARI/GP can find rational solutions of pell-like equations, if they exist. But apparantly there is no command for the integral points in Magma. Maybe, PARI/GP gives an integral solution , if it exists, but I am not sure. – Peter Apr 12 '17 at 17:55

1 Answers1

2

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There is a difference between rational solutions and integral solutions, reflected in the form class numbers.

With discriminant $101,$ we have just one class of forms up to $SL_2 \mathbb Z$ equivalence. The Gauss-Lagrange reduced form is $x^2 + 9 xy - 5 y^2.$ The numbers primitively represented are, up to $300,$ these. Note that, since $10^2 - 101 = -1,$ a number $n$ is represented if and only if $-n$ is represented.

Primitively represented positive integers up to  300

           1 =  1 
           5 = 5
          13 = 13
          17 = 17
          19 = 19
          23 = 23
          25 = 5^2
          31 = 31
          37 = 37
          43 = 43
          47 = 47
          65 = 5 * 13
          71 = 71
          79 = 79
          85 = 5 * 17
          95 = 5 * 19
          97 = 97
         101 = 101
         107 = 107
         115 = 5 * 23
         125 = 5^3
         131 = 131
         137 = 137
         155 = 5 * 31
         157 = 157
         169 = 13^2
         179 = 179
         181 = 181
         185 = 5 * 37
         193 = 193
         197 = 197
         211 = 211
         215 = 5 * 43
         221 = 13 * 17
         223 = 223
         227 = 227
         233 = 233
         235 = 5 * 47
         239 = 239
         247 = 13 * 19
         251 = 251
         281 = 281
         283 = 283
         289 = 17^2
         299 = 13 * 23


 Primitively represented positive integers up to  300

           1           9          -5   original form 

========================================================

or discriminant $404,$ there are three classes.

404    factored   2^2 *  101

    1.             1          20          -1   cycle length             2
    2.             4          18          -5   cycle length             6
    3.             5          18          -4   cycle length             6

  form class number is   3

============================================================

The first one, your $x^2 - 101 y^2,$ does not represent $\pm 71$

Primitively represented positive integers up to 300

       1 =  1 
      20 = 2^2 * 5
      37 = 37
      43 = 43
      52 = 2^2 * 13
      65 = 5 * 13
      68 = 2^2 * 17
      76 = 2^2 * 19
      85 = 5 * 17
      92 = 2^2 * 23
      95 = 5 * 19
      97 = 97
     100 = 2^2 * 5^2
     101 = 101
     115 = 5 * 23
     124 = 2^2 * 31
     125 = 5^3
     155 = 5 * 31
     179 = 179
     188 = 2^2 * 47
     221 = 13 * 17
     223 = 223
     233 = 233
     235 = 5 * 47
     247 = 13 * 19
     260 = 2^2 * 5 * 13
     283 = 283
     284 = 2^2 * 71
     299 = 13 * 23

Primitively represented positive integers up to 300

       1          20          -1   original form 

==========================================================

However, $4 x^2 + 18 xy - 5 y^2$ does

Primitively represented positive integers up to  300

           4 = 2^2
           5 = 5
          13 = 13
          17 = 17
          19 = 19
          20 = 2^2 * 5
          23 = 23
          25 = 5^2
          31 = 31
          47 = 47
          52 = 2^2 * 13
          65 = 5 * 13
          68 = 2^2 * 17
          71 = 71
          76 = 2^2 * 19
          79 = 79
          85 = 5 * 17
          92 = 2^2 * 23
          95 = 5 * 19
         100 = 2^2 * 5^2
         107 = 107
         115 = 5 * 23
         124 = 2^2 * 31
         131 = 131
         137 = 137
         148 = 2^2 * 37
         155 = 5 * 31
         157 = 157
         169 = 13^2
         172 = 2^2 * 43
         181 = 181
         185 = 5 * 37
         188 = 2^2 * 47
         193 = 193
         197 = 197
         211 = 211
         215 = 5 * 43
         221 = 13 * 17
         227 = 227
         235 = 5 * 47
         239 = 239
         247 = 13 * 19
         251 = 251
         260 = 2^2 * 5 * 13
         281 = 281
         284 = 2^2 * 71
         289 = 17^2
         299 = 13 * 23


 Primitively represented positive integers up to  300

           4          18          -5   original form 

==================================

Will Jagy
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  • @Peter you will notice that I write my own programs. The one above are based on Conway's "topograph" method, which definitively gives those numbers that are *primitively* represented. This was introduced – Will Jagy Apr 12 '17 at 19:25
  • @Peter fine. Given a rational solution, stereographic projection around that point gives the other rational solution. There is nothing about the process that allows you to correctly infer integral solutions. Meanwhile, it would appear that your main concern is software that is guaranteed to work. Mine works. I recommend you learn enough to write your own. – Will Jagy Apr 12 '17 at 19:36
  • I would like to, but it is hard to find useful informations. For example, mathworld only explains the case $|c|<\sqrt{D}$ in the pell-like equation $x^2-Dy^2=c$ and for the case $|c|>\sqrt{D}$, it is only mentioned that the solution is substantially more difficult. Neither did I find the methods in wikipedia. Where did you find the informations necessary to write your program ? – Peter Apr 12 '17 at 20:03
  • I read books. My favorite for (binary) quadratic forms is Buell. Conway's method is introduced in his Sensual Quadratic Forms. Some interesting ideas did not become clear to me until I read the short treatment in Stillwell. Neither book gives much discussion of the (elementary but important) idea of automorphism group of an indefinite form. I will paste in links to answers i have given on the topograph method... – Will Jagy Apr 12 '17 at 20:12