Exercise 5.4.9 (Equation $x^3 + 2 y^3 = 7 (u^3 + 2v^3)$)

Proof. Assume for the sake of contradiction that

Let $d=x\wedge y\wedge u\wedge v$. Then $d\ne 0$, so there are integers $X,Y,U,V$ such that

and simplifying by $d^3$,

Reducing modulo $7$, we obtain

The cubes in $\mathbb{Z}∕7\mathbb{Z}$ are $\bar{0},\bar{2},\bar{6}$, thus $-\bar{2}=\bar{5}$ is not a cube in $\mathbb{Z}∕7\mathbb{Z}$. If $Y\not\equiv 0(\mathit{mod}7)$, then $-\bar{2}={(\bar{X}{\bar{Y}}^{-1})}^3$, which is impossible since $-\bar{2}$ is not a cube. So $Y\equiv 0(\mathit{mod}7)$, thus $7\mid X^3$ and $7\mid X$. In conclusion,

Therefore $7^3\mid X^3+2Y^3=7(U^3+2V^3)$, so $7^2\mid U^3+2V^3$, which gives

By the same reasoning, $7\mid U$ and $7\mid V$, thus

This contradiction $7\mid 1$ shows that the equation $x^3+2y^3=7(u^3+2v^3)$ has no nontrivial integral solution. □