Diophantine equation of type $ax^2+bx+cy^2=n$

Question

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 $0$.

Answer 1

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.