Solutions for $65x^2-57y^2=8 \cdot 61$

Question

I am looking for solutions of $65x^2-57y^2=8 \cdot 61$.

I think there are none but I am unable to prove this. Rewriting the equation to $$57(x+y)(x-y)=8(61-x^2)$$ implies (among other things) that $57 \mid x^2-4 = (x+2)(x-2)$ and $x,y$ have the same parity, but I am unable to progress.

Answer 1

not sure how much to say. Your form obviously integrally represents $8 = 65-57.$ Your form is integrally equivalent to the "reduced" form $-7 x^2 + 120x y + 15 y^2.$

It is automatic that the prime $61$ is represented by some form of this discriminant, as $$ ( 14820 | 61) = (58| 61) = (-3|61) = (3|61) = (61|3) = (1|3) = 1. $$ The part that is fortunate is that the form that represents $61$ is the principal form $x^2 + 120 xy - 105 y^2$

It follows that we may use Dirichlet's definition of composition to produce a representation of $8 \cdot 61 = 488$ by your form.

  x :  1385     y :  1479
  x :  11227     y :  11989
  x :  42197     y :  45061
  x :  342055     y :  365271

Not sure I will work out the next specifics: there are infinitely many solutions. Given any solution, write it as a column vector, multiply a certain 2 by 2 matrix by it, you get another solution. The entries of the matrix are distressingly large and come, after some fiddling, from $$ 126202751^2 - 3705 \cdot 2073360^2 = 1 $$ Well, why not, $$ \left( \begin{array}{cc} 126202751 & 118181520 \\ 134768400 & 126202751 \end{array} \right) $$

I set the solution search program going. If I can find all solutions with $x$ up to the bound $126202751$ that will mean we have all solutions, as we can multiply each by the matrix (or its inverse). I should add an early note of optimism, as these forms do not represent $2 \mod 4$ or $4 \pmod 8$ primitively, I would expect there to be four solutions to generate the whole set, and we have already found four....Yep, we have them all. The others just need to be found by matrix multiplications.