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...
In your question, using mod will help you.
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 $7$}).$
Hence, we know that there is no solution.
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To solve this kind of equations, we have several 'tools' such as
using mod, using inequalities, using factorization...
In your question, using mod will help you.
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 $7$}).$
Hence, we know that there is no solution.