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Is there a way of solving the following equation, in integers $(x,y)$, by hand? :

$2^{3x}+17=y^2$.

You can also try: $2^{2x}+17=y^2$ or more generally $2^x+17=y^2$; each of these has at least 1 solution.

I used the elliptic curve $x^3+17=y^2$ to make this, since it had 3 solutions where $x$ is a power of 2. These are:

$(x,y)\in \{(2,\pm5), (4,\pm9), (8,\pm23)\}$.

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    $\begingroup$ Are you asking a question here, or are you just giving us orders? $\endgroup$ – Old John Aug 5 '14 at 1:00
  • $\begingroup$ It seems likely you know more about the problem than was stated. Such "questions" make it especially hard to guess what your real motivation is in asking, and discourage the Reader from making any lengthy response that might be redundant (or beyond your level of interest). $\endgroup$ – hardmath Aug 5 '14 at 1:31
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    $\begingroup$ Have you tried subtracting $16$ from both sides, then factoring them ? $\endgroup$ – Lucian Aug 5 '14 at 1:32
  • $\begingroup$ Perhaps you could add the "3 solutions where $x$ is a power of 2", referring to the last statement of your Question (using $x$ in a different sense than in the rest of the body and title). $\endgroup$ – hardmath Aug 5 '14 at 11:10
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$2^x+17=y^2$ leads to the three elliptic curves $y^2=u^3+17$, $y^2=2u^3+17$, and $y^2=4u^3+17$ (depending on $x$ modulo $3$). There are standard techniques for solving these, that is, for finding all the integral solutions. Indeed, you can write each one as a Mordell equation, $Y^2=U^3+k$; for example, from $y^2=4u^3+17$ we get $16y^2=64u^3+272$, so $Y^2=U^3+272$. Solutions to Mordell equations for various ranges of values of $k$ have been tabulated, and it should be easy to find the tables by searching the web for "Mordell equation".

$2^{2x}+17=y^2$ is easy. It's $y^2-r^2=17$, which has only the solutions $y=\pm9$, $r=\pm8$, so $x=3$.

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  • $\begingroup$ Sure there is. It says on that page there is one known solution, $x=8$. $\endgroup$ – Gerry Myerson Sep 10 '15 at 23:15

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