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I've got stuck at this problem, I don't know how to approach it.

Prove that the equation doesn't have any integer solutions for $$x^2 + y^5 = 2015^{17}$$

I've thought about Fermat's little theorem but it didn't helped. Just a hint would be really appreciated. Thanks!

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Hint: try it mod $11$.........

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  • $\begingroup$ Man ... now I have to figure out why that's a good hint ... $\endgroup$ – John Feb 16 '17 at 22:38
  • $\begingroup$ You are not alone @John dude $\endgroup$ – janmarqz Feb 17 '17 at 2:22
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    $\begingroup$ Based on your hint, this is how I did: $2015^{17} = 2015^{11} * 2015^6$. $2015^{11} \equiv 2015 \equiv 2 \pmod n$ (fermat). Then, verifying the rests mod $11$ for the first 11 squares we get $\{0, 1, 3, 4, 5, 9\}$. So, $y\equiv \{2, 3, 4, 6, 7, 9\}\pmod 5$ - impossible because $y^5\equiv \{1,10\}\pmod {11} $. $\endgroup$ – scummy Feb 17 '17 at 8:27
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    $\begingroup$ But how did you got the idea to do it mod $11$? $\endgroup$ – scummy Feb 17 '17 at 8:31
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    $\begingroup$ It's often worthwhile trying problems such as this mod small primes. $11$ has a chance because (thanks to Fermat) $y^{10}$ can only be $0$ or $1$ mod $11$, and thus $y^{5}$ only $0, 1, 10$. $\endgroup$ – Robert Israel Feb 17 '17 at 16:42

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