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I need to solve the following

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

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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.

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    $\begingroup$ Maybe an overkill, but for laziness: by FLT $y^6\equiv 1 \implies y^3\equiv\pm 1 \pmod 7$ $\endgroup$ – chubakueno Jan 6 '14 at 3:25
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    $\begingroup$ A good point. Thanks. $\endgroup$ – mathlove Jan 6 '14 at 3:26
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    $\begingroup$ @chubakueno for a brief moment, i thought the L stood for last... $\endgroup$ – Lost1 Jan 6 '14 at 4:58

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