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?