Last time I got stuck in this problem which I have posted earlier. Today I have come accross to this new situation.

How to solve the diophantine equation $ax^2+hxy+by^2+c=0$ in integers ? Given all of $a,b,c,h\in \mathbb Z$.

The motivation was to solve $30x^2+21y^2-57xy+729=0$ in integers which through MAPLE i got as $(x,y)=(-22,-17), (-10,-11), (10,11), (22,17)$.

  • $\begingroup$ also solutions to +57xy is you change 1 of the signs. $\endgroup$ – JMP Feb 5 '15 at 12:42
  • $\begingroup$ sorry i didn't get you properly $\endgroup$ – Anjan3 Feb 5 '15 at 13:33
  • 1
    $\begingroup$ Solutions to (30,57,21,729) are (22,-17) etc..., $\endgroup$ – JMP Feb 5 '15 at 13:38
  • $\begingroup$ Ohh now i got it. never though that. thanks for the info $\endgroup$ – Anjan3 Feb 6 '15 at 4:50

Here's the method to solve such an equation ( both theory and equation solver ).


| cite | improve this answer | |

In general, given an initial solution $x_0,y_0 = m,n$ to,

$$a x^2 + b x y + c y^2 + d = 0\tag1$$

if the discriminant $D= b^2-4ac\,$ is a non-square $D>0$, then an infinite more can be found as,

$$x = mu^2 - 2(b m + 2c n)u v + D m v^2$$

$$y = n u^2 + 2(2a m + b n)u v + D n v^2$$

where $u,v$ solve the Pell equation,

$$u^2 - D v^2 = \pm1\tag2$$

However, if your $D$ is a square, or $D<0$, then $(1)$ will only have a finite number of integer solutions. (In your case, it is a square $D=27^2$.)

| cite | improve this answer | |
  • $\begingroup$ Really impressive. But would you like to share with me how did you arrive such calculation like this ? $\endgroup$ – Anjan3 Feb 6 '15 at 6:33
  • $\begingroup$ @AnjanDebnath: The general method can be found in Dickson's History of the Theory of Numbers. I derived this version to highlight the connection to Pell equations. It's quite tedious to explain it step-by-step, but it's a useful identity. The more complete version can be found as Identity 2 of this post and will shed light on why it works. $\endgroup$ – Tito Piezas III Feb 6 '15 at 7:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.