Quadratic Diophantine equation $x^2+6y^2-xy=47$ has no solutions. I am trying to show that $x^2 + 6y^2 - xy = 47$ has no integer solutions. I know that the an efficient way is to look at this equation modulo $n$; other equations can be easily be solved this way. I tried this for $n = 2,3,4,5,6$ so far and I still cannot conclude that no solutions exist. Is there an efficient way of knowing what $n$ to try? Can you give some ideas for $n$ not large? Thanks.

Let $z=x/y$. Since the roots of $z^2-z+6$ are $\frac{1\pm i\sqrt{23}}2$, we get
$$
z^2-z+6=\left(z-\frac{1+i\sqrt{23}}2\right)\left(z-\frac{1-i\sqrt{23}}2\right)
$$
Multiplying by $y^2$ gives
$$
\begin{align}
47
&=x^2-xy+6y^2\\[9pt]
&=\left(x-\frac{1+i\sqrt{23}}2\,y\right)\left(x-\frac{1-i\sqrt{23}}2\,y\right)\\
&=\left(x-\frac12\,y\right)^2+\left(\frac{\sqrt{23}}2\,y\right)^2
\end{align}
$$
Multiplying by $4$ yields
$$
x^2+6y^2-xy=47\iff(2x-y)^2+23y^2=188
$$
Since $23\cdot3^2=207\gt188$, the only choices for $y$ are $\{0,\pm1,\pm2\}$.
$$
\begin{array}{r|c}
y&188-23y^2\\\hline
0&188\\
\pm1&165\\
\pm2&96
\end{array}
$$
None of the numbers in the right column are perfect squares, so none can be $(2x-y)^2$.
