Max Alekseyev on Mon, 14 Oct 2013 05:26:27 +0200

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Hi Charles,

Your equation is equivalent to
(2y+1)^2 - 8x^2 = 8n+1
and you can use bnfisintnorm(bnfinit(z^2-8),8*n+1) to get basis
solutions for (2y+1) and x as coefficients (up to a sign) of z^0 and
z^1, respectively.

E.g., for n=5, this gives [z + 7, 4*z + 13], the first one of which in
particular gives (x,y) = (1,3).


On Sun, Oct 13, 2013 at 11:02 PM, Charles Greathouse
<> wrote:
> I haven't dabbled with GP's algebraic functions very much. I recently came
> across this problem: for which n is n plus a perfect square a triangular
> number? That is, given an even number 2n, does the Diophantine equation
> y^2 + y - 2x^2 = 2n
> have a solution?
> I've only used qfbsolve and thue before, which of course don't apply here,
> but I'm pretty sure this is implemented in PARI. Hint, anyone?
> Charles Greathouse
> Analyst/Programmer
> Case Western Reserve University