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Prove that the equation $x^7 + 6x - 14y^7 = 3 $ has no solutions over $\mathbb{Z}$.

This problem is from a problem set in my Number Theory class. We discussed linear diophantine equations, how to know when solution exists and how to find all solutions after you find one solution. This 7th degree diophantine equation though is waaay above my head and I'm not sure how to even start proving the lack of solutions. Hints are appreciated. Thanks.

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$$x^7+6x-14y^7\equiv x^7-x\equiv0\pmod7$$ by Fermat's little theorem

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    $\begingroup$ +1 - Your answer came in just $7$ seconds ahead of mine! This is somewhat ironic considering the solutiong involves using modulo $7$. $\endgroup$ – John Omielan Oct 12 '19 at 4:38
  • $\begingroup$ I don't see how that follows from Fermat's little theorem. Would you please elaborate? $\endgroup$ – Sigma Oct 12 '19 at 4:47
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    $\begingroup$ @VictorS. $x^p \equiv x \pmod p$ for all $x$ and primes $p$. Thus, $x^7 \equiv x \pmod 7$. $\endgroup$ – John Omielan Oct 12 '19 at 4:49
  • $\begingroup$ @JohnOmielan The problem is that I don't see how he/she went from the first congruence to the second congruence. $\endgroup$ – Sigma Oct 12 '19 at 14:57
  • $\begingroup$ @VictorS. $$6\equiv-1\pmod7, 14\equiv0$$ $\endgroup$ – lab bhattacharjee Oct 12 '19 at 15:02

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