Find all $x,y\in\mathbb{Z}$ s.t $2x^3-7y^3=3$ Find all $$x,y\in\mathbb{Z}$$ such that $$2x^3-7y^3=3$$ 
Solution:
We consider first $$2x^3-7y^3\equiv3 \pmod 2$$ $$5y^3\equiv 1 \pmod 2$$ $$y^3\equiv 1 \pmod2$$ which has  solution  $y\equiv 1 \pmod 2$
Consider $$2x^3\equiv 3 \pmod 7$$ $$4\cdot 2x^3\equiv 4\cdot3 \pmod 7$$
$$x^3\equiv 5 \pmod 7$$
but none of $x=0,1,2,3,4,5,6\pmod 7$ satisfies the equation 
Which would mean that the equation has no integer solutions. Can this be correct?

Indeed, the equation $2x^3 - 7y^3 = 3$ has no integer solutions. Modulo $7$, we get the congruence
$$2x^3 \equiv 3 \pmod{7}$$
which every solution of the original equation must satisfy. But the group of nonzero remainders modulo $7$ has order $6$, so $x^3 \equiv \pm 1 \pmod{7}$ for all $x\not\equiv 0 \pmod{7}$. Thus, whatever $x$ is, the remainder of $x^3$ modulo $7$ is one of $-1,0,1$, and so $2x^3$ has remainder $0,2$, or $5$, so $$2x^3\not\equiv 3 \pmod{7}$$
for all $x\in\mathbb{Z}$.
