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We have an equation $x^2+4y^2-2xy-2x-4y-8=0$. Find all integer pairs $(x,y)$ satisfying this equation. I did some research on my own, and found that the above equation describes an ellipse. But I'm not sure how it helps.

Is there any systematic way to solve this?

John Griffin
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user406287
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2 Answers2

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You can write your equation as $$(x-y-1)^2+3(y-1)^2=12.$$ That means $|y-1|\le\sqrt{12/3}=2$, it's rather straightforward to check those few values.

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Hint: as an equation in $x$ the reduced discriminant of $x^2-2x(y+1)+4y^2-4y-8=0$ is:

$$ \frac{1}{4} \Delta = (y+1)^2-(4y^2-4y-8)=-3y^2+6y+9=-3(y+1)(y-3) $$

For integer solutions, the reduced discriminant must be a perfect square.

dxiv
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  • Can this method be somehow extended to quadratic equations in 3 variables? 4 variables? And above? – user406287 Aug 24 '17 at 16:08
  • @MainakRoy You can always consider the equation to be a quadratic in one of the variables, with coefficients in terms of the other variables. Then you can calculate the discriminant which will be a function of the other variables, and try to find when it's a perfect square. In the given case, that was easy because the discriminant had an obvious factorization. In the general case it may or may not be as easy. – dxiv Aug 24 '17 at 17:01