Integer solution to $19x^3-84y^2=1984$

Question

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.

Answer 1

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.

Answer 2

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 $x^3 = \pm 1 \mod 7$

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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 $y$.

Answer 3

${\rm mod}\ 7\!:\,\ 3\equiv -2x^3\,\overset{\rm square}\Rightarrow\, 3^2\equiv 4x^6\overset{\rm Fermat}\equiv 4\,\Rightarrow\!\Leftarrow$