Peter wrote: Let $ A, B, C, D, E$ be integers, $ B \neq 0$ and $ F = AD^{2} - BCD + B^{2}E \neq 0$. Prove that the number $ N$ of pairs of integers $ (x, y)$ such that \[ Ax^{2} + Bxy + Cx + Dy + E = 0, \] satisfies $ N \le 2 d( \vert F \vert )$, where $ d(n)$ denotes the number of positive divisors of positive integer $ n$. $ \Leftrightarrow Bx+D|Ax^2+Cx+E$ $ \Rightarrow Bx+D|ABx^2+BCx+BE$ $ \Rightarrow Bx+D|(Ax+C)(Bx+D)-(ADx+CD-BE)$ $ \Rightarrow Bx+D|ADx+CD-BE$ $ \Rightarrow Bx+D|ABDx+BCD-B^2E$ $ \Rightarrow Bx+D|AD(Bx+D)-AD^2+BCD-B^2E$ $ \Rightarrow Bx+D|AD^2-BCD+B^2E=F$ Imply that $ Bx+D|F$ So has at most $ 2d(|F|)$ satisfy the equation.

To be ahead of the questions: from his second last line, you have at most $ d(|F|)$ solutions for $ x$, and since the equation is quadratic, at most two $ y$ per $ x$, so at most $ 2d(|F|)$ pairs indeed.

but cant bx+d be a negative divisor as well for negative values of x? so wouldnt the answer be 4d(|F|)?