Modular arithmetic on $ 2x^3 - 7y^3 = 3 $

Question

Find integers x and y such that $2x^3-7y^3=3$

How to do it? My first thought was to reduce it to one variable problem by taking suitable mod.

1) mod 7

$ 2x^3\equiv 3 \bmod 7 $

2) mod 2

$ y^3\equiv 3 \bmod 2 $

How to solve it?

Answer 1

Solving the simplified problems should be easy:

First of all, if $y^3 \equiv 3 \equiv 1 \pmod 2$, then $y=1\pmod 2$.

If $2x^3\equiv 3 \pmod 7$, then $$ 4\cdot 2x^3\equiv 4\cdot 3 \pmod 7 $$ which means that $x^3 \equiv 5 \pmod 7$, which means that there are no solutions for $x$, since the only cubic residues modulo $7$ are $0,1$ and $6$.

So, we've proven that there are no solutions $(x,y)$ to your equation.