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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?

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  • $\begingroup$ 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. $\endgroup$ – Will Jagy Sep 26 '14 at 18:13
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    $\begingroup$ 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 $D = b^2-4ac$. $\endgroup$ – Tito Piezas III Dec 2 '14 at 16:27
  • $\begingroup$ @TitoPiezasIII: Thanks! I do prefer the original, though, as it doesn't obscure the original variables. =) $\endgroup$ – Kieren MacMillan Dec 2 '14 at 17:45
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    $\begingroup$ The general case actually yields two Pell equations. See this post. Also, a related post. $\endgroup$ – Tito Piezas III Mar 31 '16 at 20:05
  • $\begingroup$ @TitoPiezasIII: If you post an answer, I'll accept it and get this question off the Unanswered queue. $\endgroup$ – Kieren MacMillan Mar 6 '18 at 19:43

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