Prove $19m^2+95mn+2000n^2=1995$ has no integer solution for $n$ and $m$ The question read "show that $19m^2+95mn+2000n^2=1995$ has no integer solution for $n$ and $m$."
I have attempted a solution and would like to check if it is correct.
$95mn +2000n^2 = 1995-19m^2$ now factorize both sides $5n(19m+400n) = 19(105-m^2)$
From here I tried all parity cases for $m$ and $n$ and constantly end up with $odd = even$ or $even = odd$.
Is this correct? 
Thanks.
 A: Clearly, $19\mid n,n=19r$(say)
So, $$19m^2+95mn+2000n^2=1995\implies 19m^2+95m(19r)+2000(19r)^2=1995$$
So, $$m^2+95mr+2000(19)r^2=105\implies m^2\equiv10\pmod {19} $$
But $10$ is not a Quadratic residue of $19$ 
as $(\pm1)^2\equiv1\pmod{19},(\pm2)^2\equiv4,(\pm3)^2\equiv9,$
$(\pm4)^2\equiv16,(\pm5)^2=25\equiv6,(\pm6)^2=36\equiv17,$
$(\pm7)^2=49\equiv11,(\pm8)^2=64\equiv7,(\pm9)^2=81\equiv5$
A: Taking $\pmod 5$ gives $5 \mid m$. Now taking $\pmod {25}$ gives $25\mid 1995$, a contradiction.
A: If $19m^2+95mn+2000n^2 = 1995$ then reducing mod $5$ tells us that $m^2 \equiv 0 \bmod 5$, i.e. that $5|m$.
But then reducing both sides mod $25$ gives $0 \equiv 20 \bmod 25$ which is a contradiction.
A: Show that $5\mid m$ therefore $25\mid 19m^2+95mn+2000n^2$. Since $25\nmid 1995$ the result follows.
A: $\rm\begin{eqnarray}{\bf Hint}\ \ \  (p,a)\color{#C00}=1,\ \, and\, \ \ \color{#0A0}{squarefree}\  \ p\mid ax^2+bpx + cp^2\\
\rm \Rightarrow p\mid ax^2\ \color{#C00}\Rightarrow\ \rm p\mid x^2\ \color{#0A0}{\Rightarrow}\ \rm p\mid x\ \Rightarrow\ p^2\mid ax^2+bpx + cp^2\end{eqnarray}$
