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Given a general quadratic diophantine equation of the form $$ax^2+bxy+cy^2+dx+ey+f=0$$ I am interested in the sufficient conditions for the existence of a recurrence relation of the form $$X_{n+1} = PX_n+QY_n + K$$ $$ Y_{n+1} = RX_n + SY_n +L$$ that can generate all positive integer solutions $(X_n, Y_n)$ starting from some fundamental solution $(X_0, Y_0)$. I have seen the proof that such a recurrence relation exists in the case of a Pell equation $x^2-dy^2=1$ when $d$ is not a perfect square, but for the more general case I have not seen such a proof, but rather an assumption that the recurrence relation exists.

My understanding is that a necessary set of conditions for the existence of such a recurrence relation is $a)$ that the discriminant $b^2-4ac>0$ and $b)$ that there are infinitely many positive integer solutions. My specific questions are:

$0)$ Are conditions $a)$ and $b)$ indeed necessary? $b)$ obviously is, but what about $a)$?

$1)$ Are the two conditions $a)$ and $b)$ sufficient for the existence of a recurrence relation as above? If so, could someone point me towards a proof? If not, could someone provide me with a counter-example?

$2)$ Whether this set of conditions is sufficient or not, is there some "nicer" sufficient set of conditions for the existence of such a recurrence relation? In particular, $(i)$ conditions expressed only in terms of a relation between the coefficients $a, b, c, d, e, f$ of the equation, or alternatively $(ii)$ a weaker version of $b)$, for instance that there exists at least one positive integer solution.

I am specifically interested in the sufficient conditions for the existence of the recurrence relation, and not for instance in how to find the fundamental solution $(X_0, Y_0)$, nor how to compute the coefficients $P, Q, K, R, S, L$.

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    You could express your general quadratic Diophantine equation as $$(2ax+by+d)^2=(b^2-4ac)y^2+(2bd-4ae)y+(d^2-4af).$$ Do you see the problem if $b^2-4ac\leq0$? – Servaes Nov 03 '23 at 00:27
  • @Servaes I see that if $b^2-4ac<0$ then $y$ is bounded above as the RHS cannot be negative, and since $y$ is bounded the RHS is bounded, and so the LHS is bounded, so $x$ is bounded, meaning there cannot be infinitely many solutions. If $b^2-4ac=0$, then the LHS grows faster than the RHS with $y$ so $y$ is bounded above, and with the same reasoning $x$ is bounded, so there cannot be infinitely many solutions. This answers my first question, thank you! – jordisson123 Nov 03 '23 at 10:45
  • A related post about the general quadratic equation. – Tito Piezas III Jan 11 '25 at 13:41

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