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Solve the equation $x^3 + 117y^3 = 5$ over the integers.

I have tried solving this. It is clear that one of $x$ or $y$ must be negative. $117$ seemed a strange number. So I found out that $117 = 125 - 8 = 5^3 - 2^3$. I don't know if this is useful but still I'm adding it. So the equation becomes:

$$x^3 + (5y)^3 - (2y)^3 = 5$$

I don't know how to proceed further. I need some hints. Any help would be appreciated.

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    $\begingroup$ Take $\pmod{3^2}$ $\endgroup$ – lab bhattacharjee Mar 30 '16 at 8:57
  • $\begingroup$ @labbhattacharjee How did you know we had to take a $\pmod 9$? $\endgroup$ – TheRandomGuy Mar 30 '16 at 9:20
  • $\begingroup$ @Dhrub, As the power of $x,y$ are $3$ $\endgroup$ – lab bhattacharjee Mar 30 '16 at 9:32
  • $\begingroup$ @Dhruv By Binomial theorem $(3k\pm 1)^3\equiv \pm 1\pmod 9$. Similarly, $(pk+a)^p\equiv a^p\pmod{p^2}$ for any prime $p$, which gives that there are at most $p$ $p$'th powers mod $p^2$ (in this case, there are only $3$ $3$'th powers mod $3^2$, so using mod $3^2$ makes sense). $\endgroup$ – user236182 Mar 30 '16 at 14:56
  • $\begingroup$ Using mod $7$ would also make sense, because $x^3\equiv \{0,\pm 1\}\pmod{7}$, but unfortunately it doesn't solve it. By Fermat's Little Theorem $x^6\equiv \{0,1\}\pmod{7}$, so $x^3\equiv \{0,\pm 1\}\pmod{7}$. More generally, $x^{\frac{p-1}{2}}\equiv \{0,\pm 1\}\pmod{p}$ for any odd prime $p$ (see Euler's Criterion for a stronger result, namely $x^{\frac{p-1}{2}}\equiv \left(\frac{x}{p}\right)\pmod{p}$ for any odd prime $p$). $\endgroup$ – user236182 Mar 30 '16 at 15:17
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Hint: $$x^3 \equiv 0,1,8 (\bmod 9)$$ $$117y^3 \equiv 0 (\bmod 9)$$ $$5 \equiv 5 (\bmod 9)$$

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  • $\begingroup$ Admittedly, that is nearly the full proof, but +1. $\endgroup$ – S.C.B. Mar 30 '16 at 8:58
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    $\begingroup$ Haha, and here I was asking Sage to calculate the class group of $\mathbb Q(\sqrt[3]{-117})$. +1 $\endgroup$ – RKD Mar 30 '16 at 8:58
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    $\begingroup$ @Dhruv: I saw that $9|117$ and remembered that $x^3 \equiv 0,1,8 (\bmod 9) $ $\endgroup$ – Roman83 Mar 30 '16 at 10:08
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    $\begingroup$ @Roman83 Great thinking. Thanks. $\endgroup$ – TheRandomGuy Mar 30 '16 at 10:08
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    $\begingroup$ To prove that $x^3\equiv \{0,\pm 1\}\pmod{9}$, notice that by Binomial Theorem it's easy to see that $(3k\pm 1)^3\equiv \pm 1\pmod{9}$. $\endgroup$ – user236182 Mar 30 '16 at 14:48

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