Chapter 21.2 of Friedlander and Iwaniec's *Opera de Cribro* begins (essentially) as follows:


Let
$$F(X, Y) = aX^2 + bXY + cY^2 + \alpha X + \beta Y + \gamma \in \mathbb{Z}[X, Y]$$ be irreducible in $\mathbb{Q}[X, Y]$, represent arbitrarily large odd integers, and satisfy $(a, b, c, \alpha, \beta, \gamma) = 1$. We want $F(X, Y)$ to depend essentially on two variables, which may be expressed by the requirement that $\frac{\partial F}{\partial X}, \frac{\partial F}{\partial Y}$ are linearly independent.


We introduce two discriminants:
$$\Delta = b^2 - 4 ac,$$
$$D = a \beta^2 - b \alpha \beta + c \alpha^2 + \Delta \gamma.$$


[end of excerpt]


Wait, they do what? 


F+I go on to state that any $F(X, Y)$ satisfying the above conditions represents infinitely many primes: on the order of $\frac{x}{\log x}$ primes $\leq x$ if $D = 0$ or $\Delta$ is a square, and on the order of $\frac{x}{(\log x)^{3/2}}$ otherwise.


They say only a very little bit about their proof, referring instead to [this](http://matwbn.icm.edu.pl/ksiazki/aa/aa24/aa2451.pdf) paper of Iwaniec. Obviously these discriminants are important (Iwaniec calls them the "small" and "large" discriminants respectively), but Iwaniec does not explain them in any highbrow fashion; he simply dives into the guts of some computations. Predictably, they are related to affine changes of coordinates (we want to say that $F, F'$ are equivalent if an invertible such transformation changes one into the other), but if the role of $D$ is something simple then this wasn't apparent from a quick look at the paper.


Are these two discriminants classical and familiar? Is there a simple highbrow explanation of what they say about the polynomial $F$? It seems there must be.


(See also [here](https://mathoverflow.net/questions/55384/primes-represented-by-two-variable-quadratic-polynomials) for a related MO question.)


Thank you!



'Small' and 'large' discriminants of two-variable quadratic polynomials
Here are some naive comments. If $X, Y$ are large then the contribution of the quadratic terms swamps the other terms, so first let's concentrate on the quadratic terms
$$Q(X, Y) = a X^2 + b XY + c Y^2.$$


This can be written as
$$Q(X, Y) = \left[ \begin{array}{cc} X & Y \end{array} \right] \left[ \begin{array}{cc} a & \frac{b}{2} \\\ \frac{b}{2} & c \end{array} \right] \left[ \begin{array}{c} X \\\ Y \end{array} \right]$$


and the matrix occurring above has determinant $ac - \frac{b^2}{4} = - \frac{\Delta}{4}$. $Q$ is positive-definite if and only if $\Delta < 0$, in which case this determinant has concrete geometric significance: it is $\pi$ times the reciprocal of the area of the ellipse $Q(X, Y) \le 1$. Consequently it describes the asymptotics of the number $q\_n$ of pairs of integers $(X, Y)$ such that $Q(X, Y) \le n$ (and the same should be true of $F$). In all cases, $\Delta$ is invariant under affine change of coordinates. 


$F$ itself can be written as


$$F(X, Y) = \left[ \begin{array}{ccc} X & Y & 1 \end{array} \right] \left[ \begin{array}{ccc} a & \frac{b}{2} & \frac{\alpha}{2} \\\
 \frac{b}{2} & c & \frac{\beta}{2} \\\
 \frac{\alpha}{2} & \frac{\beta}{2} & \gamma \end{array} \right] \left[ \begin{array}{c} X \\\ Y \\\ 1 \end{array} \right]$$


and $D$ is $4$ times the determinant of this matrix. Again this is invariant under affine change of coordinates. $D$ is a natural invariant of the homogenization of $F$ to a ternary quadratic form. 




---


Let me also say some naive things about the two conditions appearing in the OP. By the quadratic formula, the quadratic polynomial $ax^2 + bx + c$ has roots
$$\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.$$


Consequently, $\Delta$ is a square if and only if this polynomial has rational roots, hence if and only if $Q$ splits into a product of factors 
$$Q(X, Y) = (dX + eY)(fX + gY)$$


over $\mathbb{Z}$ (by Gauss' lemma). As Will Jagy says, these quadratic forms have very different behavior as far as representing primes compared to quadratic forms that do not factor this way.


If $\Delta$ is not a square, then it is in particular not zero. Now, if $D = 0$, then the $3 \times 3$ matrix above has nontrivial nullspace. I believe this is equivalent to being able to write $F$ as the product of two linear polynomials (after a suitable quadratic extension) as opposed to the product of two linear polynomials plus a constant. 


