Is there a recipe for, or are there practical examples of, solving Diophantine equations of type $ax^2+bx+cy^2=n$. How would I prove that a particular equation has no ( Integer ) solutions for $x, y$? $(a, b, c, n)$ are integers not equal to $0$.
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For $b=0$ and $c<0$ see Pell's Equation. I think your question is quite broad. – Rory Daulton Mar 18 '15 at 10:41
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Ok, b can't be zero: I'll edit the question. ( I know how to solve Pell equations using continued fraction tables. ) – nilo de roock Mar 18 '15 at 11:15
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@ndrook1: Your question is essentially about a quadratic equation to be made a multiple of a square $ax^2+bx-n = -cy^2$. This has been well-studied. If there is an initial integer solution and $ac$ is neither a square nor positive, then an infinite more can be found using a Pell equation. A good applet is the Alpertron. – Tito Piezas III Mar 19 '15 at 04:00
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Why don't you make an answer of this? – nilo de roock Mar 21 '15 at 12:11
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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}
Setting $b=e=0$ and $f=-n$, and then rewriting $d \to b$, we have your equation $$ ax^2+bx+cy^2=n. $$
There have also been myriad solutions of special cases in the past couple hundred years. An excellent survey and reference is Dickson’s History of the Theory of Numbers, with Volumes II and III of particular interest.
Kieren MacMillan
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