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Find all triples $(x,y,z)$ of positive integers such that $$2018^x=y^2+z^2+1$$

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closed as off-topic by GNUSupporter 8964民主女神 地下教會, Did, Namaste, user223391, erfink Apr 28 '18 at 19:19

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  • $\begingroup$ Immediately $x$ must be odd $\endgroup$ – lab bhattacharjee Apr 27 '18 at 9:44
  • $\begingroup$ Why $2018{}{}$? $\endgroup$ – Lord Shark the Unknown Apr 27 '18 at 9:45
  • $\begingroup$ Seems like some math contest stuff... $\endgroup$ – Thern Apr 27 '18 at 9:47
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    $\begingroup$ HINT: Look at this equation modulo $4$. You can easily see that necessarily $x<2$. Then the equation becomes $$z^2+y^2=2017$$ which has only two symmetric solutions $(9,44)$ or $(44,9)$. $\endgroup$ – Crostul Apr 27 '18 at 9:53
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HINT: Look at this equation modulo $4$. You can easily see that necessarily $x<2$ (i.e. $x=1$).

Then the equation becomes $$z^2+y^2=2017$$ which has only two symmetric solutions $(9,44)$ or $(44,9)$.

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