<s>[INST] 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.
 [/INST] 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 $y$.
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