0
$\begingroup$

I'm having problems with finding all possible integers solutions of particular equations, like this one for example: $x^2 -xy + 2y^2 = 29$. What sets me off, is the term $xy$, I don't know how to deal with it.

I know methods for solving equations like $x^2 - 3y^2 = -3$ (then I work with the fundamental unit and with norm forms), but I'm stuck on a method for the first one. Substitutions don't seem to do the trick, but I could be wrong.

As always, any help would be dearly appreciated.

$\endgroup$
  • 1
    $\begingroup$ You can multiply by $4$ and complete the square to obtain $4x^2-4xy+8y^2=(2x-y)^2+7y^2=116$ and set $y=2z$ since y has to be even to get $(x-z)^2+7z^2=29$ $\endgroup$ – Mark Bennet Aug 24 '15 at 13:40
  • $\begingroup$ Oh yes, that's a way to do it in the case $p=29$, thanks! Do you know of some general method for these cases? I can imagine that those tricks don't always work or that you don't always see it (ahem). $\endgroup$ – Riley Aug 24 '15 at 13:44
  • $\begingroup$ @Riley By "in general" do you mean when $p$ is arbitrary? Or even more arbitrary like $$ax^2-bxy+cy^2 = p$$ $\endgroup$ – graydad Aug 24 '15 at 13:47
  • $\begingroup$ $4a(ax^2+bxy+cy^2)=(2ax+by)^2+(4ac-b^2)y^2$ so there is always a way of completing the square. $\endgroup$ – Mark Bennet Aug 24 '15 at 13:48
  • $\begingroup$ @graydad, never mind, Mark Bennet has already answered my question. :) So if I understand it correctly: there's always a way to rewrite the equation so that you can apply the methods as mentioned in the question? $\endgroup$ – Riley Aug 24 '15 at 13:52
1
$\begingroup$

$$x=\dfrac{y\pm\sqrt{y^2-4(2y^2-29)}}2=\dfrac{y\pm\sqrt{116-7y^2}}2$$

We need $116-7y^2$ to be $\ge0$ and perfect sqaure

$y^2\le\dfrac{116}7<17$

So, we need to test for $1\le y\le4$

$\endgroup$

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.