0
$\begingroup$

I’m trying to solve the Diophantine equation \begin{align} \tag{$\star$} (2^k-3)b^2 - (2^{k+1}+1)ab - (2^k+2)a^2 = 0, \end{align} where $k$ is a positive integer, and $a$ and $b$ are relatively prime integers (not necessarily positive). A computer search turns up only the solutions $k=1$ and $k=4$, with no more solutions $1 \le k \le 100$.

I had hoped that I could use Vieta-jumping to solve it, but after much trying and many sheets of sketch paper, I haven’t found the magic incantation. Any help — either with a Vieta-jumping hint, or with any hints toward a proof mechanism — would be greatly appreciated.

In case it helps, I know that $$ (b^2-2ab-a^2) \mid 2 \cdot 23. $$ Also, it is trivial to show that ($\star$) implies $2^{2k+3}=c^2+23$ for some integer $c$.

$\endgroup$
  • 1
    $\begingroup$ It has an integer solution if and only if the discriminant is a square $\endgroup$ – Will Jagy Aug 23 '16 at 0:43
  • $\begingroup$ Yes (cf. my edit of 11 minutes ago). The question then is, can ($\star$) be attacked in any other way? $\endgroup$ – Kieren MacMillan Aug 23 '16 at 0:47
1
$\begingroup$

You have a minor variant of Ramanujan Nagell. I would imagine a complete resolution is available somewhere.

https://en.wikipedia.org/wiki/Ramanujan%E2%80%93Nagell_equation

jagy@phobeusjunior:~$ ./Pell_Target_Fundamental  Automorphism matrix:  
    3   8
    1   3
  Automorphism backwards:  
    3   -8
    -1   3

  3^2 - 8 1^2 = 1

 x^2 - 8 y^2 = -23

Mon Aug 22 18:00:32 PDT 2016

x:  3  y:  2 ratio: 1.5  SEED   KEEP +-   y =   2
x:  7  y:  3 ratio: 2.33333  SEED   BACK ONE STEP  -3 ,  2  y =   3
x:  25  y:  9 ratio: 2.77778  y =   3^2
x:  45  y:  16 ratio: 2.8125  y =   2^4
x:  147  y:  52 ratio: 2.82692  y =   2^2 13
x:  263  y:  93 ratio: 2.82796  y =   3 31
x:  857  y:  303 ratio: 2.82838  y =   3 101
x:  1533  y:  542 ratio: 2.82841  y =   2 271
x:  4995  y:  1766 ratio: 2.82843  y =   2 883
x:  8935  y:  3159 ratio: 2.82843  y =   3^5 13
x:  29113  y:  10293 ratio: 2.82843  y =   3 47 73
x:  52077  y:  18412 ratio: 2.82843  y =   2^2 4603
x:  169683  y:  59992 ratio: 2.82843  y =   2^3 7499
x:  303527  y:  107313 ratio: 2.82843  y =   3 35771
x:  988985  y:  349659 ratio: 2.82843  y =   3^2 38851
x:  1769085  y:  625466 ratio: 2.82843  y =   2 277 1129
x:  5764227  y:  2037962 ratio: 2.82843  y =   2 1018981
x:  10310983  y:  3645483 ratio: 2.82843  y =   3 1215161
x:  33596377  y:  11878113 ratio: 2.82843  y =   3 13 151 2017
x:  60096813  y:  21247432 ratio: 2.82843  y =   2^3 2655929

Mon Aug 22 18:02:32 PDT 2016

 x^2 - 8 y^2 = -23

jagy@phobeusjunior:~$ 
$\endgroup$
  • $\begingroup$ Based on your observation, I found Apery's theorem that equations of the form $$x^2+D=2^n$$ have at most two solutions. I'll look at that to see what I can see. Still going to try to solve ($\star$) directly/independently, though… because, based on my derivation, an effective attack could be applied to a lot more equations than just the Ramanujan-Nagell type. $\endgroup$ – Kieren MacMillan Aug 23 '16 at 1:33
  • 1
    $\begingroup$ @KierenMacMillan $x^2 + 7 = 2^n$ has about five solutions, so Apery's theorem cannot be quite what you wrote. $\endgroup$ – Will Jagy Aug 23 '16 at 1:37
  • $\begingroup$ Yes, $D=7$ is the single exception (q.v. en.wikipedia.org/wiki/…). $\endgroup$ – Kieren MacMillan Aug 23 '16 at 1:40
  • 1
    $\begingroup$ @KierenMacMillan this math.tifr.res.in/~saradha/saradharev.pdf says $x^2 + 23 = 2^n$ has the ones we know along with $45^2 + 23 = 2048 = 2^{11}$ I see, that is my $y=16.$ $\endgroup$ – Will Jagy Aug 23 '16 at 1:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.