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

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.

• Take $\pmod{3^2}$ – lab bhattacharjee Mar 30 '16 at 8:57
• @labbhattacharjee How did you know we had to take a $\pmod 9$? – TheRandomGuy Mar 30 '16 at 9:20
• @Dhrub, As the power of $x,y$ are $3$ – lab bhattacharjee Mar 30 '16 at 9:32
• @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). – user236182 Mar 30 '16 at 14:56
• 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 $p$). – user236182 Mar 30 '16 at 15:17

Hint: $$x^3 \equiv 0,1,8 (\bmod 9)$$ $$117y^3 \equiv 0 (\bmod 9)$$ $$5 \equiv 5 (\bmod 9)$$
• Haha, and here I was asking Sage to calculate the class group of $\mathbb Q(\sqrt[3]{-117})$. +1 – RKD Mar 30 '16 at 8:58
• @Dhruv: I saw that $9|117$ and remembered that $x^3 \equiv 0,1,8 (\bmod 9)$ – Roman83 Mar 30 '16 at 10:08
• To prove that $x^3\equiv \{0,\pm 1\}\pmod{9}$, notice that by Binomial Theorem it's easy to see that $(3k\pm 1)^3\equiv \pm 1\pmod{9}$. – user236182 Mar 30 '16 at 14:48