<s>[INST] Integer solutions to $5m^2-6mn+7n^2 = 1985$ 
Are there integers $m$ and $n$ such that $$5m^2-6mn+7n^2 = 1985?$$

Taking the equation modulo $3$ gives $n^2-m^2 \equiv 2 \pmod{3}$. Thus, $3 \mid n$ but $3 \nmid m$. How can I use this to find a contradiction?
 [/INST] The discriminant of the quadratic form is $6^2-4\cdot 5\cdot 7=-2^3\cdot \color{blue}{13}$ and the equation is equivalent to:
$$  (5m-3n)^2+26 n^2 = 9925 \tag{1}$$
that gives:
$$ 6\equiv 9925 \equiv (5m-3n)^2 \pmod{\color{blue}{13}}\tag{2} $$
but $6$ is not a quadratic residue $\!\!\pmod{13}$, since $13\equiv 5\pmod{8}$ and $13\equiv 1\pmod{3}$, so:
$$ \left(\frac{6}{13}\right)=\left(\frac{-1}{13}\right)\left(\frac{2}{13}\right)\left(\frac{-3}{13}\right)=(+1)\cdot(-1)\cdot(+1)=\color{red}{-1}\tag{3}$$
and there are no integer solutions to the original equation.
For similar problems, quadratic reciprocity and prime factors of the discriminant are the key.
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