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?

Taking modulo $5$ we have
$$n(2n-m)\equiv 0\pmod 5$$
This gives us two possibilities:


*

*If $n$ is a multiple of $5$, write $n=5u$ and $175u^2-30mu+5m^2=1985$, or $35u^2-6mu+m^2=397$. Then


$$m=\frac{6u\pm\sqrt{1588-104u^2}}{2}=3u\pm\sqrt{397-26u^2}$$


*

*If $2n\equiv m\pmod 5$, then $m=5u+2n$ and (after some easy calculations) $m^2-14mu-29u^2+397=0$. Now,
$$m=\frac{14u\pm\sqrt{312u^2-1588}}2=7u\pm\sqrt{78u^2-397}$$


Both possibilities require $-397$ to be a square modulo $13$, but $-397\equiv -7\pmod{13}$, and $-7$ is not a square modulo $13$.
Thus, there is no solution.
