# Solve $7x^3+2=y^3$ over integers

I need to solve the following

solve $7 x^3 + 2 = y^3$ over integers. How can I do that?

To solve this kind of equations, we have several 'tools' such as

using mod, using inequalities, using factorization...

Since we have $$y^3-2=7x^3,$$ the following has to be satisfied : $$y^3\equiv 2\ \ \ (\text{mod 7}).$$
However, in mod $7$, $$0^3\equiv 0,$$ $$1^3\equiv 1,$$ $$2^3\equiv 1,$$ $$3^3\equiv 6,$$ $$4^3\equiv 1,$$ $$5^3\equiv 6,$$ $$6^3\equiv 6.$$
So, there is no integer $y$ such that $y^3\equiv 2\ \ \ (\text{mod$7$}).$
• Maybe an overkill, but for laziness: by FLT $y^6\equiv 1 \implies y^3\equiv\pm 1 \pmod 7$ – chubakueno Jan 6 '14 at 3:25