Integer solution to $19x^3-84y^2=1984$ Show that there exist no integer values $x,y$ such that $19x^3-84y^2=1984$.
Please help me in understanding no solution problems.
I tried to check the modulo $7$ of both sides but couldn't reject some cases.
 A: Working modulo 7,
$$
19x^3 − 84y^2 \equiv 1984 \\
5x^3 \equiv 3 \\
15x^3 \equiv 9 \\
x^3 \equiv 2 \\
$$
By inspection, $x^3 \equiv 2 \mod 7 \,$ has no solutions:
$$
\begin{array}{c|r|c} 
x & x^3 & x^3\mod 7 \\
\hline
0 & 0 & 0 \\
1 & 1 & 1 \\
2 & 8 & 1 \\
3 & 27 & 6 \\
4 & 64 & 1 \\
5 & 125 & 6 \\
6 & 216 & 6 \\
\end{array}
$$
Or we could invoke Fermat's little theorem and note that for all $x: x\mod 7 \neq 0,\, x^6 = 1$, hence $x^3 = \pm 1 \mod 7$

Alternatively, working mod 19
$$
19x^3 − 84y^2 \equiv 1984 \\
 − 8y^2 \equiv 8 \\
$$
Note that $7 \times 8 = 56 \equiv -1 \mod 19$
So
$$y^2 \equiv -1 \mod 19$$
But according to the the first supplement to the law of quadratic reciprocity
$$y^2 \equiv -1 \mod p$$ for prime $p$ only has solutions if $p  \equiv 1 \mod 4 $ . And since $19 \mod 4 \equiv 3\,$ there are no solutions for $y$.
A: Rewrite the equation as $19(x^3-100)$$=84(1+y^2)$. Now check the remainders when divided by $7$. The RHS gives $0$, so LHS should also be a multiple of $7$. This implies  $$ 7 \mid (x^3-100)$$$$\implies 7 \mid (x^3-2)$$ and Voila! No cube of a natural number gives remainder $2$ when divided by $7$. Proof: I know a rigorous and tedious one. If somebody knows a shorter one, please enlighten me. Consider the number to be of the form $7k+n$, where $k=0,1\cdots$ and $n \in [0,6]$. Do the cube and you just need to find the remainder when $n^3$ is divided by $7$. I think it is pretty clear from here .And thus, it completes the proof that no integral solutions exist.
A: ${\rm mod}\ 7\!:\,\ 3\equiv -2x^3\,\overset{\rm square}\Rightarrow\, 3^2\equiv 4x^6\overset{\rm Fermat}\equiv 4\,\Rightarrow\!\Leftarrow$
