Hi Would you please advise me? Consider the equation below: $$ ax^2+bxy+cy^2=n $$ in which $a, b, c$ and $n$ are integers. We then suppose that $a, b, c$ are constant. Is there any way to find the number of answers for the equation? Actually, I have already solved this equation for many different $a, b, c$ and the number of answers. And currently I'm looking for unsolved cases. By special thanks
2026-02-28 01:15:26.1772241326
A question about integral quadratic forms
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Assume that $\Delta = b^2 - 4 a c$ is equal to $d$, if $\frac{d-1}{4}$ is an integer, or equal to $4 d$, where $\frac{d - 2}{4}$ or $\frac{d - 3}{4}$ is an integer, and in both cases assume that $d$ is not divisible by the square of any integer. If for every odd prime divisor $p$ of $n$, you can find a non-negative integer $t$ less than $p$ such that $\frac{\Delta - t^2}{p}$ is an integer, and if $n$ is even, then $\frac{\Delta \pm 1}{8}$ is an integer, then there exist integers $x, y$ satisfying $a x^2 + b x y + c y^2 = n$. If at least one solution $x, y$ exists then the number of solutions depends on whether $\Delta < 0$, in which case there are only finitely many. If $\Delta > 0$ and at least one solution exists, then there are infinitely many. If $\Delta < 0$, then the precise number should be proportional to $\sum_{m \mid n} (\frac{\Delta}{m})$, where $(\frac{\Delta}{m})$ is the Kronecker symbol, http://en.wikipedia.org/wiki/Kronecker_symbol, and we are adding over each divisor $m$ of $n$.