Given that $\left(\frac{10}{23}\right)=-1$. How would I go about showing that $9x^2-46(y^3+3y+1)=10$ has no integer solutions?
I believe it has something to do with Quadratic Reciprocity.
For example, we can show that $x^2-43y^2=73$ has no integer solutions by demonstrating that $x^2 \equiv 73(mod\:43)$ has no solutions as $$\left(\frac{73}{43}\right)=\left(\frac{30}{43}\right)=\left(\frac{2}{43}\right)\left(\frac{3}{43}\right)\left(\frac{5}{43}\right)=\left(\frac{43}{5}\right)=\left(\frac{3}{5}\right)=-1$$
I also believe that this shows that $x^2-43y^n=73, n \in Z$ has no integer solutions.
I imagine I am missing some jump between $9x^2-46(y^3+3y+1)=10$ and a Legendre Symbol that can be transformed into $\left(\frac{10}{23}\right)$.
Hint: Taking mod 23, we get that $ (3x)^2 \equiv 10 \pmod{23}$.
What does $\left( \frac{10}{23} \right) = - 1$ tell us about existence of solutions $x$?