# Solving Quadratic Diophantine Equation with initial solutions.

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!
• Do you know about Pell equations? In all these cases you could complete the square and multiply through by a constant to obtain a Pell equation. You can then use the method in your link. Hope this helps. – Zestylemonzi Sep 22 '15 at 14:38
• Led same formula which reduces this equation to the equation Pell. She's not to like? – individ Sep 22 '15 at 14:44
• @Zestylemonzi could you please show me how you can convert my equation to Pell equation then use the method in my link to get the soltuions? Put this as an answer and I will consider it as an answer if it DOES NOT use factorization in it procedure. – Zrieq Sep 22 '15 at 14:51

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.

• @n-s Thanks for this solution, I already know it. I am asking about a way to generate integers solutions using the initial solutions that I already have. if you read the link in the question, you will find it is using an initial solution to get another bunch of other integer solutions. – Zrieq Sep 22 '15 at 14:45
• @Zrieq The method you point to is coming from the Pell equation, and generates INFINITELY many solutions, so cannot work for an equation which has finitely many solutions. It is possible in theory to find a recurrence which leads to finitely many solutions, but it cannot be something similar to the Pell solution. – N. S. Sep 22 '15 at 14:52
• @N-S I am looking for a recurrence formula or something that benefits from existing solutions, I don't need to solve Pell's Equation and end factoring some numbers check this to understand the story. – Zrieq Sep 22 '15 at 16:15