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Does the equation

$$y^2 = 2x^2 - n$$

where $n$ is an integer have many rational $(x,y)$ solutions?

We can always express $n$ as

$$n = a^2-b^2$$

for some $a$ and $b$. For example, $a=(n+1)/2$ and $b=(n-1)/2$. Then if $a^2+b^2 =c^2$, we have $(x,y)=(a,b)$ as one solution. Is there a systematic way of finding other solutions?

Scene
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Jim
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    If you have a solution of $a^2-2b^2=1$ and a solution of your equation, then multiplying both equations and using Brahmagupta's identity you get new solutions of your equation. – plop Jan 10 '23 at 21:44
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    Jim, I do not follow how your $n=a^2-b^2,a^2+b^2=c^2$ can give you $b^2=2a^2-n.$ – Anne Bauval Jan 10 '23 at 21:54
  • Meanwhile: always $ (x,y) \mapsto (3x+4y, 2x+3y)$ gives a new solutions from old, and applieds arbitrarily often fives infinitely many. If $n = \pm 1$ there is a single "seed" solution, either $(1,0)$ or $(1,1),$ all others derive from it. $n = \pm 2$ is a special case....For $n$ being any prime $p \equiv \pm 1 \pmod 8$ there are two seed solutions, $(x_0, \pm y_0) $ For $n4 being a product of such primes, the number of seeds goes up (and is a power of two). – Will Jagy Jan 10 '23 at 21:59
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    @AnneBauval I used to give answers (on indefinite binary quadratic forms) with Conway's Topograph, which shows nicely how the solutions in between images under the form "automorph" fit in. However, in some ten years, I never convinced any student to draw such a thing. – Will Jagy Jan 10 '23 at 22:16
  • Thank you for all your suggestions. I should have been clearer in stating the problem. What I really want to know is whether we can find rational solutions to the equation 2=22− for a fixed value of n where n is a positive or a negative integer. – Jim Jan 11 '23 at 21:02

2 Answers2

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well, why not. $x^2 - 2 y^2 = 2737 = 7 \cdot 17 \cdot 23$ needs eight seed solutions to show all solutions with $x,y \geq 0.$ Because $2^3 = 8$

To add eight to the row number, take that $(x,y)$ and apply $(x,y) \mapsto (3x+4y, 2x+3y).$ Example, row 1 says $(53,6),$ so row 9 should be $(3 \cdot 53 + 4 \cdot 6, 2 \cdot 53 + 3 \cdot 6) = (183, 124)$

$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$

1. x:  53      y: 6 =  2 3  SEED   KEEP +- 
2. x:  55      y: 12 =  2^2 3  SEED   KEEP +- 
3. x:  57      y: 16 =  2^4  SEED   KEEP +- 
4. x:  73      y: 36 =  2^2 3^2  SEED   KEEP +- 
5. x:  75      y: 38 =  2 19  SEED   BACK ONE STEP  73 ,  -36
6. x:  107      y: 66 =  2 3 11  SEED   BACK ONE STEP  57 ,  -16
7. x:  117      y: 74 =  2 37  SEED   BACK ONE STEP  55 ,  -12
8. x:  135      y: 88 =  2^3 11  SEED   BACK ONE STEP  53 ,  -6
9. x:  183      y: 124 =  2^2 31
10. x:  213      y: 146 =  2 73
11. x:  235      y: 162 =  2 3^4
12. x:  363      y: 254 =  2 127
13. x:  377      y: 264 =  2^3 3 11
14. x:  585      y: 412 =  2^2 103
15. x:  647      y: 456 =  2^3 3 19
16. x:  757      y: 534 =  2 3 89
17. x:  1045      y: 738 =  2 3^2 41
18. x:  1223      y: 864 =  2^5 3^3
19. x:  1353      y: 956 =  2^2 239
20. x:  2105      y: 1488 =  2^4 3 31
21. x:  2187      y: 1546 =  2 773
22. x:  3403      y: 2406 =  2 3 401
23. x:  3765      y: 2662 =  2 11^3
24. x:  4407      y: 3116 =  2^2 19 41

$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$

 w^2 - 2 v^2 = -2737 =   -1 * 7 17 23

Tue 10 Jan 2023 02:17:00 PM PST


    

  1. x:  1      y: 37 =  37  SEED   KEEP +-
    2. 

  2. x:  25      y: 41 =  41  SEED   KEEP +-
    3. 

  3. x:  31      y: 43 =  43  SEED   KEEP +-
    4. 

  4. x:  41      y: 47 =  47  SEED   KEEP +-
    5. 

  5. x:  65      y: 59 =  59  SEED   BACK ONE STEP  -41 ,  47
    6. 

  6. x:  79      y: 67 =  67  SEED   BACK ONE STEP  -31 ,  43
    7. 

  7. x:  89      y: 73 =  73  SEED   BACK ONE STEP  -25 ,  41
    8. 

  8. x:  145      y: 109 =  109  SEED   BACK ONE STEP  -1 ,  37
    9. 

  9. x:  151      y: 113 =  113
    10. 

  10. x:  239      y: 173 =  173
    11. 

  11. x:  265      y: 191 =  191
    12. 

  12. x:  311      y: 223 =  223
    13. 

  13. x:  431      y: 307 =  307
    14. 

  14. x:  505      y: 359 =  359
    15. 

  15. x:  559      y: 397 =  397
    16. 

  16. x:  871      y: 617 =  617
    17. 

  17. x:  905      y: 641 =  641
    18. 

  18. x:  1409      y: 997 =  997
    19. 

  19. x:  1559      y: 1103 =  1103
    20. 

  20. x:  1825      y: 1291 =  1291
    21. 

  21. x:  2521      y: 1783 =  1783
    22. 

  22. x:  2951      y: 2087 =  2087
    23. 

  23. x:  3265      y: 2309 =  2309
    24. 

  24. x:  5081      y: 3593 =  3593
    25.

$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$

Will Jagy
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I'll assume that you asked for rational solutions (whereas Diophantine equation means integer solutions).

Let $N(u+v\sqrt2 )=u^2-2v^2$.

First consider the rational parametrization of $N(a+b\sqrt{2})=1$ which is

$$(a,b)=(\frac{1+2 t^2}{1-2t^2},\frac{2 t}{1-2t^2})$$

Then we need to find one rational solution to $N(y+x\sqrt{2})=-n$, all the other solutions will be $$N((\frac{1+2 t^2}{1-2t^2}+\frac{2 t^2}{1-2t^2}\sqrt{2})(y+x\sqrt{2}))=-n$$

$N(1+\sqrt2)=-1$ so a rational solution to $N(y+x\sqrt{2})=-n$ exists iff a rational solution to $|N(y+x\sqrt{2})|=|n|$ exists.

From the splitting of prime numbers in the PID $\Bbb{Z}[\sqrt2]$ and quadratic reciprocity we get that a solution to $|N(y+x\sqrt{2})|=|n|$ exists iff $|n|=2^a\prod_j p_j^{e_j}$ with each $p_j^{e_j}\equiv \pm 1\bmod 8$.

reuns
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