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During some free time I had, I was wondering how to find the integer solutions $(x,y,z)$ to this generalized equation: $$z^2=axy+bx+cy+d$$ I am specifically looking for ways that do no involve factoring. And $a,b,c,d$ are all non-zero integers. I have no idea if it is easier or harder that for solving in two variables.

Edit: I have done some research and have concluded that it is a two-sheeted hyperboloid. I don't know if this helps with solving my question.

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  • $\begingroup$ Solving Diophantine equations without factoring? Good luck! $\endgroup$ – Gerry Myerson May 5 at 4:17
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$$z^2=axy+bx+cy+d$$ Use another equation. $$q=\frac{A^2-d}{b}$$

And we use solutions to the Pell equation. $k,t -$ any number.

$$p^2-akts^2=1$$

Decisions then write down so.

$$z=Ap^2-((aq+c)t+bk)ps+aAkts^2$$

$$x=qp^2-2kAps+(k((aq+c)t+bk)-aqkt)s^2$$

$$y=ts(((aq+c)t+bk)s-2Ap)$$

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  • $\begingroup$ Can you give a specific example and solve. And the way you wrote your answer makes it seem that there are infinite solutions, when I know that is not true. But still, very good. +1 $\endgroup$ – Quote Dave May 5 at 19:13
  • $\begingroup$ For anyone $A$ you can choose an infinite number of coefficients. The Pell equation has an infinite number of solutions... $\endgroup$ – individ May 6 at 5:12
  • $\begingroup$ So how would you solve an equation like this: $z^2=2xy-6y-12x+3$? $\endgroup$ – Quote Dave May 6 at 15:06
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Change the odds.... write it down....

$$z^2=2xy-6x-12y+3$$

$$z^2=axy+bx+cy+d$$

Let's pick some numbers...... like this. $k=3$ ; $t=2$ ; $A=3$

$$q=\frac{A^2-d}{b}=-1$$

$$p^2-2*3*2s^2=p^2-12s^2=1$$

Knowing the first solution.... $p=7$ ; $s=2$

The rest find on formula..

$$p_2=7p+24s$$

$$s_2=2p+7s$$

Solutions then write using the formula....

$$z=3p^2+46ps+36s^2$$

$$x=-(p^2+18ps+126s^2)$$

$$y=-4s(23s+3p)$$

$p=97$ ; $s=28$

$z=181387$

$x=-157081$

$y=-104720$

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  • $\begingroup$ That is not what Wolfram says: wolframalpha.com/input/?i=z%5E2%3D2xy-6y-12x%2B3 $\endgroup$ – Quote Dave May 6 at 20:03
  • $\begingroup$ Wolfram says there are 3 solutions only. But in fact, they are infinitely many. Choose other coefficients and you will receive those decisions which are mentioned there. $\endgroup$ – individ May 7 at 4:12
  • $\begingroup$ Have you checked the solutions you got to make sure they are right, and for what coefficients do you get those answers? $\endgroup$ – Quote Dave May 7 at 15:14
  • $\begingroup$ Also, Wolfram says you're wrong for the solution you got: wolframalpha.com/input/… $\endgroup$ – Quote Dave May 7 at 15:17
  • $\begingroup$ You need to be careful when carrying out arithmetic operations.... You made mistakes. I won't answer any more arithmetic questions.... $\endgroup$ – individ May 7 at 16:34
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Above equation shown below:

$z^2=axy+bx+cy+d$ -----------$(1)$

Equation (1) has parametric solution for $(a,b,c,d)= (3,9,2,25)$

By the way "Quote Dave" has made a mistake in claiming

that numerical solution given by "Individ" is incorrect.

The solution $(x,y,z)=(-157081,-104720,181387)$ satisfies

equation $(1)$ for $(a,b,c,d)=(2,-6,-12,3)$.

For, $(a,b,c,d)= (3,9,2,25)$ the solution is:

$x=[(k^2-14k+20)/(k^2-3)]$

$y=[(3k^2-14k+14)/(k^2-3)]$

$z=[(7k^2-23k+21)/(k^2-3)]$

For $k=2$, we get:

$(x,y,z)=(-4,-2,3)$

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