I have read here about a method to generate integer solutions for a Diophantine equations like the following:
\begin{align*}
an^2+bn+c = d^2
\end{align*}
By knowing an initial integer solution for it.
Now I am wondering if there is a method to get the integer solutions for the following Quadratic Diophantine equation:
\begin{align*}
y^2+y-x^2-bx+c=0
\end{align*}
by knowing an initial integer solutions for this equation.
For example, for the following equation I know the first two initial integer solutions which are $x_1=47$ & $x_2=19$
\begin{align*}
y^2+y-x^2-21x+4=0
\end{align*}
So is there a method like the one mentioned in the another link such that I can find any other integer solutions using these initial solutions?
Notes:
- What is considered as a solution for this questions is to use the existing known solutions to get other integer solutions (if exists) by some recurrence formula OR some method that benefits from the existing solutions.
- if there is no known way to solve this, then let's start thinking of some way to build a relationship between initial integer solutions and the rest of the solutions.
- Please don't think of factorization as a part of the new solution at all because this will not help. The reason why it doesn't help could be found in the notes of this questions.
- I know that solving Pell's Equation for this question will give me all integer solutions. So I will not consider any such answers as a solution because it uses factorization at the end of the day.
- Please don't vote up an answer that doesn't meet the question needs, Thanks!
Best Answer
Multiplying with 4 (I first completed the square and then multiplied to get rid of fractions), the equation becomes $$(2y+1)^2-(2x+21)^2+456=0$$
Thus $$(2x+21)^2-(2y+1)^2=456$$ Factor the LHS $$(2x-2y+20)(2x+2y+22)=456\\ (x-y+10)(x+y+11)=114$$
Now factor $114=2*3*19$ in all the $24$ possible ways each of them gives you exactly one solution.
P.S. The solution $x=19, y=27$ corresponds to $$2 * 57=114$$ The solution $x=19, y=-28$ corresponds to $$57 * 2=114$$
P.S. In general, by this method any equation of the form \begin{align*} y^2+ay-x^2-bx+c=0 \end{align*}
Can be reduced to an equation of teh form \begin{align*} (2y+a)^2-(2x+b)^2=\alpha \end{align*} which by factoring the LHS leads to finitely many solutions.
Equations of the form \begin{align*} y^2+ay-dx^2-bx+c=0 \end{align*} can by multiplication by $4d^2$ be reduced to the Pell equation.