Quadratic Diophantine to Pell reduction

Question

http://mathworld.wolfram.com/DiophantineEquation2ndPowers.html says the equation $$ax^2+bxy+cy^2=k$$ can be reduced to Pell equation.

Can someone explain how?

Answer 1

To sketch the how, starting with $ax^2+bxy+cy^2=k$, look at it as a quadratic in $x$. We then need
$$x = \frac{-b\pm\sqrt{b^2-4a(cy^2-k)}}{2a}$$ to be an integer. Clearly this needs $\sqrt{b^2-4a(cy^2-k)} = z$ to be an integer.

So we solve $z^2 + (4ac)y^2 = b^2 + 4ak$, which is a generalised Pell equation and I suppose you are familiar with solving.

Of course we need to (in the end) pick solutions only where $2a \mid (-b \pm z)$.

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Another approach is to use a general translation $X = px + qy, Y = ry$ and then pick the values such that the cross term disappears, then complete the square etc... If I recall the specific translation, will post it. An old reference I have is http://www.math.niu.edu/~rusin/known-math/95/quadratics (check part 2)