# Is Legendre’s solution of the general quadratic equation the only one?

Legendre famously solved the general quadratic equation $$ax^2+bxy+cy^2+dx+ey+f=0$$ by noting that \begin{equation*} 4a(b^2-4ac)(ax^2+bxy+cy^2+dx+ey+f) = 0 \tag{$\star$} \end{equation*} along with the assumption $a(b^2-4ac) \ne 0$ forces $ax^2+bxy+cy^2+dx+ey+f=0$, while simultaneously implying the Pell equation \begin{align} &\bigl((b^2-4ac)y-2ae+bd\bigr)^2 - (b^2-4ac)(2ax+by+d)^2 \\ &\hspace{16em}= 4a(ae^2+b^2f+cd^2-bde-4acf). \tag{$\dagger$} \end{align}

Evidently, we could substitute for ($\star$) any equation of the form $$k(ax^2+bxy+cy^2+dx+ey+f)=0,$$ where $k$ is some nonzero constant or function or set of variables, and thence derive another equation like ($\dagger$).

My question is, what other such “complete solutions” have been found?

• this is about it. Complete the square in one variable by multiplying by $4a,$ there is still non-diagonal stuff left, so you do it again, this is the result. Finding the eigenvectors and center generally involves irrational numbers, no good for diophantine equations. Commented Sep 26, 2014 at 18:13
• You can make $\dagger$ slightly more concise as, $$(D y - 2a e + b d)^2 - D(2a x + b y + d)^2 = 4a(a e^2 + c d^2 - b d e + D f)$$ with discriminant $D = b^2-4ac$. Commented Dec 2, 2014 at 16:27
• @TitoPiezasIII: Thanks! I do prefer the original, though, as it doesn't obscure the original variables. =) Commented Dec 2, 2014 at 17:45
• The general case actually yields two Pell equations. See this post. Also, a related post. Commented Mar 31, 2016 at 20:05
• @TitoPiezasIII: If you post an answer, I'll accept it and get this question off the Unanswered queue. Commented Mar 6, 2018 at 19:43