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Find all $x,y\in \mathbb{N}$ such that: $2^x+17=y^2$.

Its easy to find that $x=6$ is the only even value for $x$, the others have to be odd. One more thing is that we get $y^2 \equiv 19 \pmod p$, for every prime factor of $x$. But I have no ides what next to do.

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    $\begingroup$ Just a little useless remark: This is an equation "of Ramanujan-Nagell type": there are some interesting properties of this kind of equations on en.wikipedia.org/wiki/Ramanujan%E2%80%93Nagell_equation; but it really doesn't help your specific problem :-) $\endgroup$
    – PseudoNeo
    Sep 8, 2015 at 18:22
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    $\begingroup$ See math.stackexchange.com/q/887651/264 $\endgroup$
    – Zev Chonoles
    Sep 8, 2015 at 18:23
  • $\begingroup$ @ZevChonoles It is a solution to the problem, but I think there shouldnt be a more elementar way to do it. $\endgroup$
    – CryoDrakon
    Sep 8, 2015 at 18:33
  • $\begingroup$ What is the source of this problem? $\endgroup$
    – user236182
    Nov 30, 2015 at 18:18
  • $\begingroup$ Why are you claiming that $2^x\equiv 2\pmod{p}$ for every prime factor $p$ of $x$? $\endgroup$
    – user236182
    Nov 30, 2015 at 18:48

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