# Integer Solutions to a Two-Sheeted Hyperboloid

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.

• Solving Diophantine equations without factoring? Good luck! – Gerry Myerson May 5 at 4:17

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

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

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

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

• 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 – Quote Dave May 5 at 19:13
• For anyone $A$ you can choose an infinite number of coefficients. The Pell equation has an infinite number of solutions... – individ May 6 at 5:12
• So how would you solve an equation like this: $z^2=2xy-6y-12x+3$? – Quote Dave May 6 at 15:06

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....

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

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

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

$$p=97$$ ; $$s=28$$

$$z=181387$$

$$x=-157081$$

$$y=-104720$$

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

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:

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